Question

The distance between electric charges Q C and 9Q C is 4 m. What is the electric potential at a point on the line joining them where the electric field is zero ?(A) 4 kQ V
(B) 10 kQ V
(C) 2 kQ V
(D) 2.5 kQ V

Answer 1

Answer:

A)4kQ V

Explanation:

Hope it help you

mark as Brainliest answer

Answer 2

The electric potential at a point is 4KQ volts.

Explanation:

Let us consider joining point P where the electric field is zero.

Electric field at point P is:

$\bold{E = \frac{K Q}{r^{2}}}$

Electric potential at point P is:

$\bold{V=\frac{K Q}{r}}$

The distance between P and charge Q is x

The distance between P and charge 9Q is 4 - x

From question, heir magnitude is equal to get zero electric field, so the direction of electric field is opposite.

$\bold{\frac{K Q}{x^{2}}=\frac{9 K Q}{(4-x)^{2}}}$

$\bold{\Rightarrow \frac{1}{x^{2}}=\frac{9}{(4-x)^{2}}}$

On taking square root, we get,

$\bold{\Rightarrow \frac{1}{x}=\frac{3}{(4-x)}}$

Now, the distance x = 1 m

The electric potential at point P is sum of Potential due charge QC and  Potential due charge 9Q.

$\bold{\Rightarrow V= \frac{K Q}{1}+\frac{9 K Q}{4-1}}$

$\bold{\Rightarrow V = \frac{K Q}{1}+\frac{9 K Q}{3}}$

$\bold{\therefore V = 4K Q \ v}$