If the electric potential along a line due to a charge is given by $V (r) = \frac{25}{r} + 3$ , where r is the distance from the charge, what is the electric field at a point r = 5 along this line?

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Solution

The correct option is A
We are given an r dependent equation for potential. So we can directly find a relation for potential by differentiating the V equation.

$V(r) = \frac{25}{r} +3$
$E (r) = - \frac{d V}{d r}$
$∴ E r = - \frac{25}{r^{2}}$
At r = 5, $E (5) = - \frac{25}{25}$
$E (5) = - 1 \frac{N}{C}$

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If the electric potential along a line due to a charge is given by V(r)=25r+3, where r is the distance from the charge, what is the electric field at a point r = 5 along this line?

The correct option is A We are given an r dependent equation for potential. So we can directly find a relation for potential by differentiating the V equation. \(V(r) = \frac{25}{r} +3\) E(r)=−dVdr ∴Er=−25r2 At r = 5, E(5)=−2525 E(5)=−1NC

If the electric potential along a line due to a charge is given by $V (r) = \frac{25}{r} + 3$ , where r is the distance from the charge, what is the electric field at a point r = 5 along this line?