Question

When charge of 3 coulomb is placed in a uniform electric field , it experiences a force of 3000 newton , within this field, potential difference between two points separated by a distance of 1 cm is?

Answer 1

Answer: The potential difference between two points is 10 volt.

Explanation

Given that,

Force F = 3000 N

Charge q = 3 coulomb

Distance d = 1 cm

Using formula

$F = q E$

$F = q\dfrac{V}{d}$

Where, F = Force

q = Charge

V = Potential difference

D = Distance between two points

$3000 =3\times\dfrac{V}{1\times10^{-2}}$

$V = 10\ volt$

Hence, The potential difference between two points is 10 volt.

Answer 2

Answer:

Potential difference, V = 10 volts

Explanation:

It is given that,

Magnitude of charge, q = 3 C

Force acting on the charge, F = 3000 N

We have to find the potential difference between two points separated by a distance of 1 cm or 0.01 m

We know that the force acting on the charge in electric field is given by :

F = q E

Electric potential, V = E × d

So, $F=\dfrac{qV}{d}$

$V=\dfrac{Fd}{q}$

$V=\dfrac{3000\ N\times 0.01\ m}{3\ C}$

V = 10 volts

Hence, this is the required solution.