Question

The electric potential at a distance of 10m from a point charge of 25 N/C

Answer 1

Answer:

Use Gauss law

$e \times 4\pi \times {10}^{2} = \frac{25}{ \epsilon} \\ e = \frac{25}{ 4\pi\epsilon} = 25 \times 9 \times {10}^{9} = 2.25 \times {10}^{11}$

Answer 2

Answer:

The electric potential at a distance of 10m from a point charge of 25nC is 22.5V

Explanation:

Given that,

The distance from point charge(r)=10m

The magnitude of point charge(Q)=25nC

According to gauss law,the potential at a distance r from a point charge of magnitude Q is given as

$V= \frac{kQ}{r}$

Here the value of k is

$k = 9 \times 10 {}^{9} Nm {}^{2} C {}^{ - 2}$

Substituting the known values in the above relation,we get potential as follows

$V = \frac{9 \times 10 {}^{9} \times 25 \times 10 {}^{ - 9} }{10} = 22.5V$

Therefore,The electric potential at a distance of 10m from a point charge of 25nC is 22.5V

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