Question

In electric field a 6.75 charge experience 2.5N force when palced at distance of 5m from origin then potential gradient at this point will be

Answer 1

Explanation:

V/R is potential gradient and answer is

2.5/6.75 = 0.37

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Answer 2

Potential gradient, $\dfrac{dV}{dl}=0.37\ V/m$

Explanation:

It is given that,

Charge, q = 6.75 C

Electric force, F = 2.5 N

Distance, d = 5 m

We need to find the potential gradient at this point. It is given by :

$\dfrac{dV}{dl}$. It is equal to E such that,

$E=\dfrac{dV}{dl}$

$E=\dfrac{F}{q}=\dfrac{dV}{dl}$

$\dfrac{dV}{dl}=\dfrac{F}{q}$

$\dfrac{dV}{dl}=\dfrac{2.5}{6.75}$

Potential gradient, $\dfrac{dV}{dl}=0.37\ V/m$

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