electric field practice problems

8/4/2019 Electric Field Practice Problems

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Electric Field Practice ProblemsSub Topics

1. Practice Problems on Electric Field Practice Problems

Introduction to electric field practice problems:-

Problem:- Force of 2.25 N is acting on charge 15*10-4 C at any point. Find electric field intensity at that

point.

Solution:- Formula for calculating electric field when force is given = force/charge = F/q

= 2.25N/15*10-4C

E = 1500 N/C (Answer)

Problem :- Calculate electric field at distance of 1 Å (Angstrom) from nucleus of Helium atom.

Solution:- Formula for calculating the field = E = (1/4πε0) (q/r2)

charge in nucleus q = 2*1.6*10-19 C

distance = r = 10-10 m

E = 9*109 Nm2 /C2 2*1.6*10-19C/(10-10)2 m2

= 28.8*10-10 /10-20 N/C

E = 2.88*1011 N/C (Answer)

Problem:- A charge density of ρL = 10PC/m is uniformly distributed along an infinite line. Determine

electric field at a point ( , , 1).

Solution:- Field due to infinite line charge = E = ρ L/2πε0ρ*aρ , where ρL = linear charge density,

ρ = distance between

linear charge and point where field has to be

determined

Field at the ( , , 1) is

http://www.tutorvista.com/physics/electric-field-practice-problems#practice-problems-on-electric-field-practice-problems





E = ρL/2πε0ρ*aρ ,

where ρL = 10PC/m, ρ = in cylindrical coordinates.

ρ = = 2,

also 1/4πε0 = 9*109 , therefore 1/2πε0 = 18* 109

E = (18* 109 *10*10-12 )/2 aρ

= 9*10-2 aρ

E = .09 aρ V/m (Answer)

a. An airplane accelerates down a runway at 3.20 m/s2 for 32.8 s until is

finally lifts off the ground. Determine the distance traveled before takeoff.

See solution below.

b. A car starts from rest and accelerates uniformly over a time of 5.21

seconds for a distance of 110 m. Determine the acceleration of the car.

See solution below.

c. Upton Chuck is riding the Giant Drop at Great America. If Upton free falls

for 2.6 seconds, what will be his final velocity and how far will he fall?

See solution below.

d. A race car accelerates uniformly from 18.5 m/s to 46.1 m/s in 2.47seconds. Determine the acceleration of the car and the distance traveled.

See solution below.

e. A feather is dropped on the moon from a height of 1.40 meters. The

acceleration of gravity on the moon is 1.67 m/s2. Determine the time for the

feather to fall to the surface of the moon.

See solution below.

f. Rocket-powered sleds are used to test the human response to acceleration.

If a rocket-powered sled is accelerated to a speed of 444 m/s in 1.8 seconds, then

what is the acceleration and what is the distance that the sled travels?

See solution below.

g. A bike accelerates uniformly from rest to a speed of 7.10 m/s over a

distance of 35.4 m. Determine the acceleration of the bike.

See solution below.

http://www.physicsclassroom.com/class/1dkin/u1l6d.cfm#sol1























h. An engineer is designing the runway for an airport. Of the planes that will

use the airport, the lowest acceleration rate is likely to be 3 m/s2. The takeoff

speed for this plane will be 65 m/s. Assuming this minimum acceleration, what is

the minimum allowed length for the runway?

See solution below.

i. A car traveling at 22.4 m/s skids to a stop in 2.55 s. Determine the

skidding distance of the car (assume uniform acceleration).

See solution below.

j. A kangaroo is capable of jumping to a height of 2.62 m. Determine the

takeoff speed of the kangaroo.

See solution below.

k. If Michael Jordan has a vertical leap of 1.29 m, then what is his takeoff

speed and his hang time (total time to move upwards to the peak and then return

to the ground)?

See solution below.

l. A bullet leaves a rifle with a muzzle velocity of 521 m/s. While accelerating

through the barrel of the rifle, the bullet moves a distance of 0.840 m. Determine

the acceleration of the bullet (assume a uniform acceleration).

See solution below.

m. A baseball is popped straight up into the air and has a hang-time of 6.25 s.

Determine the height to which the ball rises before it reaches its peak. (Hint: the

time to rise to the peak is one-half the total hang-time.)

See solution below.

n. The observation deck of tall skyscraper 370 m above the street. Determine

the time required for a penny to free fall from the deck to the street below.

See solution below.

o. A bullet is moving at a speed of 367 m/s when it embeds into a lump of

moist clay. The bullet penetrates for a distance of 0.0621 m. Determine the

acceleration of the bullet while moving into the clay. (Assume a uniform

acceleration.)

See solution below.

p. A stone is dropped into a deep well and is heard to hit the water 3.41 s

after being dropped. Determine the depth of the well.

See solution below.

q. It was once recorded that a Jaguar left skid marks that were 290 m in

length. Assuming that the Jaguar skidded to a stop with a constant acceleration of

-3.90 m/s2, determine the speed of the Jaguar before it began to skid.






































See solution below.

r. A plane has a takeoff speed of 88.3 m/s and requires 1365 m to reach that

speed. Determine the acceleration of the plane and the time required to reach this

speed.

See solution below.

s. A dragster accelerates to a speed of 112 m/s over a distance of 398 m.

Determine the acceleration (assume uniform) of the dragster.

See solution below.

t. With what speed in miles/hr (1 m/s = 2.23 mi/hr) must an object be

thrown to reach a height of 91.5 m (equivalent to one football field)? Assume

negligible air resistance.

See solution below.

Solutions to Above Problems

a.

Given:

a = +3.2 m/s2

t = 32.8 s vi = 0 m/s

Find:

d = ??

b. d = vi*t + 0.5*a*t2

c. d = (0 m/s)*(32.8 s)+ 0.5*(3.20 m/s2)*(32.8 s)2

d. d = 1720 m

e. Return to Problem 1

f.

g.

Given:

d = 110 m t = 5.21 s vi = 0 m/s

Find:

a = ??

h. d = vi*t + 0.5*a*t2

i. 110 m = (0 m/s)*(5.21 s)+ 0.5*(a)*(5.21 s)2

j. 110 m = (13.57 s2)*a

k. a = (110 m)/(13.57 s2)









http://www.physicsclassroom.com/class/1dkin/u1l6d.cfm#q1









l. a = 8.10 m/ s2

m. Return to Problem 2

n.

o.

Given:

a = -9.8 m t = 2.6 s vi = 0 m/s

Find:

d = ??

vf = ??

p. d = vi*t + 0.5*a*t2

q. d = (0 m/s)*(2.6 s)+ 0.5*(-9.8 m/s2)*(2.6 s)2

r. d = -33 m (- indicates direction)

s. vf = vi + a*t

t. vf = 0 + (-9.8 m/s2)*(2.6 s)

u. vf = -25.5 m/s (- indicates direction)

v. Return to Problem 3

w.

x.

Given:

vi = 18.5 m/s vf = 46.1 m/s t = 2.47 s

Find:

d = ??

a = ??

y. a = (Delta v)/t

z. a = (46.1 m/s - 18.5 m/s)/(2.47 s)

aa. a = 11.2 m/s2

bb. d = vi*t + 0.5*a*t2

cc. d = (18.5 m/s)*(2.47 s)+ 0.5*(11.2 m/s2)*(2.47 s)2

dd. d = 45.7 m + 34.1 m









ee. d = 79.8 m

ff. (Note: the d can also be calculated using the equation vf 2 = vi

2 + 2*a*d)

gg. Return to Problem 4

hh.

ii.

Given:

vi = 0 m/s d = -1.40 m a = -1.67 m/s2

Find:

t = ??

jj. d = vi*t + 0.5*a*t2

kk. -1.40 m = (0 m/s)*(t)+ 0.5*(-1.67 m/s2)*(t)2

ll. -1.40 m = 0+ (-0.835 m/s2)*(t)2

mm. (-1.40 m)/(-0.835 m/s2) = t2

nn. 1.68 s2 = t2

oo. t = 1.29 s

pp. Return to Problem 5

qq.

rr.

Given:

vi = 0 m/s vf = 44 m/s t = 1.80 s

Find:

a = ??

d = ??

ss. a = (Delta v)/t

tt. a = (444 m/s - 0 m/s)/(1.80 s)

uu. a = 247 m/s2

vv. d = vi*t + 0.5*a*t2

ww. d = (0 m/s)*(1.80 s)+ 0.5*(247 m/s2)*(1.80 s)2

xx. d = 0 m + 400 m









yy. d = 400 m

zz. (Note: the d can also be calculated using the equation vf 2 = vi

2 + 2*a*d)

aaa. Return to Problem 6

bbb.

ccc.

Given:

vi = 0 m/s vf = 7.10 m/s d = 35.4 m

Find:

a = ??

ddd. vf 2 = vi

2 + 2*a*d

eee. (7.10 m/s)2 = (0 m/s)2 + 2*(a)*(35.4 m)

fff. 50.4 m2 /s2 = (0 m/s)2 + (70.8 m)*a

ggg. (50.4 m2 /s2)/(70.8 m) = a

hhh. a = 0.712 m/s2

iii. Return to Problem 7

jjj.

kkk.

Given:

vi = 0 m/s vf = 65 m/s a = 3 m/s2

Find:

d = ??

lll. vf 2 = vi

2 + 2*a*d

mmm. (65 m/s)2 = (0 m/s)2 + 2*(3 m/s2)*d

nnn. 4225 m2 /s2 = (0 m/s)2 + (6 m/s2)*d

ooo. (4225 m2 /s2)/(6 m/s2) = d

ppp. d = 704 m

qqq. Return to Problem 8

rrr.

sss.












Given:

vi = 22.4 m/s vf = 0 m/s t = 2.55 s

Find:

d = ??

ttt. d = (vi + vf )/2 *t

uuu. d = (22.4 m/s + 0 m/s)/2 *2.55 s

vvv. d = (11.2 m/s)*2.55 s

www. d = 28.6 m

xxx. Return to Problem 9

yyy.

zzz.

Given:

a = -9.8 m/s2

vf = 0 m/s d = 2.62 m

Find:

vi = ??

aaaa. vf 2 = vi

2 + 2*a*d

bbbb. (0 m/s)2 = vi2 + 2*(-9.8 m/s2)*(2.62 m)

cccc. 0 m2 /s2 = vi2 - 51.35 m2 /s2

dddd. 51.35 m2 /s2 = vi2

eeee. vi = 7.17 m/s

ffff. Return to Problem 10

gggg.

hhhh.

Given:

a = -9.8 m/s2

vf = 0 m/s d = 1.29 m

Find:

vi = ??

t = ??

iiii. vf 2 = vi

2 + 2*a*d

jjjj. (0 m/s)2 = vi2 + 2*(-9.8 m/s2)*(1.29 m)

kkkk. 0 m2 /s2 = vi2 - 25.28 m2 /s2









llll. 25.28 m2 /s2 = vi2

mmmm. vi = 5.03 m/s

nnnn. To find hang time, find the time to the peak and then double it.

oooo. vf = vi + a*t

pppp. 0 m/s = 5.03 m/s + (-9.8 m/s2)*tup

qqqq. -5.03 m/s = (-9.8 m/s2)*tup

rrrr. (-5.03 m/s)/(-9.8 m/s2) = tup

ssss. tup = 0.513 s

tttt. hang time = 1.03 s

uuuu. Return to Problem 11

vvvv.

wwww.

Given:

vi = 0 m/s vf = 521 m/s d = 0.840 m

Find:

a = ??

xxxx. vf 2 = vi2 + 2*a*d

yyyy. (521 m/s)2 = (0 m/s)2 + 2*(a)*(0.840 m)

zzzz. 271441 m2 /s2 = (0 m/s)2 + (1.68 m)*a

aaaaa. (271441 m2 /s2)/(1.68 m) = a

bbbbb. a = 1.62*105 m /s2

ccccc. Return to Problem 12

ddddd.

eeeee.

Given:

a = -9.8 m/s2

vf = 0 m/s t = 3.13 s

Find:

d = ??









a. (NOTE: the time required to move to the peak of the trajectory is

one-half the total hang time.)

First use: vf = vi + a*t

0 m/s = vi + (-9.8 m/s2)*(3.13 s)

0 m/s = vi - 30.6 m/s

vi = 30.6 m/s

Now use: vf 2 = vi

2 + 2*a*d

(0 m/s)2 = (30.6 m/s)2 + 2*(-9.8 m/s2)*(d)

0 m2 /s2 = (938 m/s) + (-19.6 m/s2)*d

-938 m/s = (-19.6 m/s2)*d

(-938 m/s)/(-19.6 m/s2) = d

d = 47.9 m

Return to Problem 13

fffff.

Given:

vi = 0 m/s d = -370 m a = -9.8 m/s2

Find:

t = ??

ggggg. d = vi*t + 0.5*a*t2

hhhhh. -370 m = (0 m/s)*(t)+ 0.5*(-9.8 m/s2)*(t)2

iiiii. -370 m = 0+ (-4.9 m/s2)*(t)2

jjjjj. (-370 m)/(-4.9 m/s2) = t2

kkkkk. 75.5 s2 = t2

lllll. t = 8.69 s

mmmmm. Return to Problem 14

nnnnn.

ooooo.









Given:

vi = 367 m/s vf = 0 m/s d = 0.0621 m

Find:

a = ??

ppppp. vf 2 = vi

2 + 2*a*d

qqqqq. (0 m/s)2 = (367 m/s)2 + 2*(a)*(0.0621 m)

rrrrr. 0 m2 /s2 = (134689 m2 /s2) + (0.1242 m)*a

sssss. -134689 m2 /s2 = (0.1242 m)*a

ttttt. (-134689 m2 /s2)/(0.1242 m) = a

uuuuu. a = -1.08*106 m /s2

vvvvv. (The - sign indicates that the bullet slowed down.)

wwwww. Return to Problem 15

xxxxx.

yyyyy.

Given:

a = -9.8 m/s2

t = 3.41 s vi = 0 m/s

Find:

d = ??

zzzzz. d = vi*t + 0.5*a*t

2

aaaaaa. d = (0 m/s)*(3.41 s)+ 0.5*(-9.8 m/s2)*(3.41 s)2

bbbbbb. d = 0 m+ 0.5*(-9.8 m/s2)*(11.63 s2)

cccccc. d = -57.0 m

dddddd. (NOTE: the - sign indicates direction)

eeeeee. Return to Problem 16

ffffff.

gggggg.

Given:

a = -3.90 m/s2

vf = 0 m/s d = 290 m

Find:

vi = ??

hhhhhh. vf 2 = vi

2 + 2*a*d









iiiiii. (0 m/s)2 = vi2 + 2*(-3.90 m/s2)*(290 m)

jjjjjj. 0 m2 /s2 = vi2 - 2262 m2 /s2

kkkkkk. 2262 m2 /s2 = vi2

llllll. vi = 47.6 m /s

mmmmmm. Return to Problem 17

nnnnnn.

oooooo.

Given:

vi = 0 m/s vf = 88.3 m/s d = 1365 m

Find:

a = ??

t = ??

pppppp. vf 2 = vi

2 + 2*a*d

qqqqqq. (88.3 m/s)2 = (0 m/s)2 + 2*(a)*(1365 m)

rrrrrr. 7797 m2 /s2 = (0 m2 /s2) + (2730 m)*a

ssssss. 7797 m2 /s2 = (2730 m)*a

tttttt. (7797 m2 /s2)/(2730 m) = a

uuuuuu. a = 2.86 m/s2

vvvvvv. vf = vi + a*t

wwwwww. 88.3 m/s = 0 m/s + (2.86 m/s2)*t

xxxxxx. (88.3 m/s)/(2.86 m/s2) = t

yyyyyy. t = 30. 8 s

zzzzzz. Return to Problem 18

aaaaaaa.

bbbbbbb.

Given: Find:

a = ??











vi = 0 m/s vf = 112 m/s d = 398 m

ccccccc. vf 2 = vi

2 + 2*a*d

ddddddd. (112 m/s)2 = (0 m/s)2 + 2*(a)*(398 m)

eeeeeee. 12544 m2 /s2 = 0 m2 /s2 + (796 m)*a

fffffff. 12544 m2 /s2 = (796 m)*a

ggggggg. (12544 m2 /s2)/(796 m) = a

hhhhhhh. a = 15.8 m/s2

iiiiiii. Return to Problem 19

jjjjjjj.

kkkkkkk.

Given:

a = -9.8 m/s2

vf = 0 m/s d = 91.5 m

Find:

vi = ??

t = ??

lllllll. First, find speed in units of m/s:

mmmmmmm. vf 2 = vi

2 + 2*a*d

nnnnnnn. (0 m/s)2 = vi2 + 2*(-9.8 m/s2)*(91.5 m)

ooooooo. 0 m2 /s2 = vi2 - 1793 m2 /s2

ppppppp. 1793 m2 /s2 = vi2

qqqqqqq. vi = 42.3 m/s

rrrrrrr. Now convert from m/s to mi/hr:

sssssss. vi = 42.3 m/s * (2.23 mi/hr)/(1 m/s)

ttttttt. vi = 94.4 mi/hr




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