Question

An infinitely long line of charge has linear density of
charge equal to 1.0 nC/m. Points A and B are at a
distance of 2 m and 3 m respectively from the line of
charge. Potential difference between points A and B,
VA - Vg will be equal to​

Answer 1

Correct question: An infinitely long line of charge has linear density of charge equal to 1.0 nC/m. Points A and B are at a distance of 2 m and 3 m respectively from the line of charge. Potential difference between points A and B, $V_{A} - V_{B}$ will be equal to​:

1. 9 V

2. 10 ln3 V

3. 18 ln$\frac{3}{2}$ V

4. 9 ln $\frac{2}{3}$ V

Correct answer is option (3). 18 ln$\frac{3}{2}$ V.

Given:

Linear charge density, λ = 1.0 nC/m = $10^{-9} .$

Distance of point A from line of charge = 2m.

Distance of point B from line of charge = 3m.

To Find:

In order to find the potential difference (V) between the points A and B, we make use of the following formulas.

Electric field intensity due to line of charge, E = $\frac{2kλ}{r} .$.

Coulomb's constant, k = 9 × $10^{9}$ $Nm^{2} /C^{2} .$

Solution:

We know that,

E = $\frac{-dV}{dr} .$

From the above equation we get V as,

V =  $-\int\limits^A_B {E} \,. dr$

On applying the limits and substituting the E = $\frac{2kλ}{r} .$, the above equation becomes:

$V_{A} - V_{B}$ = $-\int\limits^2_3 {\frac{2kλ}{r} } \,. dr$

$V_{A} - V_{B}$ = $- 2$ × $9$ × $10^{9}$ × $10^{-9}$ (ln)²₃

$V_{A} - V_{B}$ = $- 18 ln \frac{2}{3}$

= $18 ln \frac{3}{2} V.$

Hence the potential difference between points A and B, $V_{A} - V_{B}$ is $18 ln \frac{3}{2} V.$

#SPJ1