Question

A point charge develops an electric potential of 10 J/C and a n

electric field of 40N/C at a point. Calculate the magnitude of the

charge and the distance from the point charge. What will happen to

the potential and field if the magnitude of the charge is doubled and

the distance is halved?​

Answer 1

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a point charge develpos an electric field pf 40N/C and a potential diffrence of 10/C at a point .calculate the magnitude of charge and distance of point charge

kajal, 3 years ago

Grade:12

1 Answers

Vikas TU

14149 Points

3 years ago

Dear Student,

Given information :

Electric Field = 40 N/C

Potential Difference = 10/C

As we realize that

Electric field = kq/r^2

Potential distinction = kq/r

Electric Field/potential distinction = kq/r^2/kq/r

40/10 = 1/r

 r = ¼ = 0.25 m

As we definitely realize that

V = kq/r

10 = 36 * 10^9 *q

 q = 10^-8/36

 q = 2 * 10^-10 C .

Cheers!!

Regards,

Answer 2

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