Question

21. What is the electric potential at a distance of 9 cm
from 3 nC?
[KCET 2014]
(a) 270 V
(b) 3 V
(c) 300 V
(d) 30 V​

Answer 1

Answer:

3 volt will be the electric potential.

Answer 2

Answer:

The correct answer is option (c) $300 V$.

Explanation:

Concept:

The amount of work required to move a unit positive test charge from infinity to a specific location inside an electric field—along any path—is known as the electric potential at that location. Electric potential is the effort required to transport a unit charge from one point to another in the presence of an electric field. Earth is typically chosen as the reference point, but any location outside the range of the electric field charge can be used.

Given:

$\quad r=9 \mathrm{~cm}$

$r$$=0.09 \mathrm{~m}$

Charge $Q=3 n C$

$Q=3 \times 10^{-9} C$

To find:

We have to find the required the electric potential.

Solution:

The required potential at a distance is,

$&9 \mathrm{~cm}$, $\quad V=\frac{k Q}{r} \quad$

Here, $k=9 \times 10^{9}$

Thus,

$V=\frac{9 \times 10^{9} \times 3 \times 10^{9}}{0.09}$

$V=300 \mathrm{~V}$

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