Question

1. The dimensions of fall of potential per unit distance are given by:
(a) [MLT A')
(b) [ML T'A')
(C) [ML T'A
(d) [MLT A​

Answer 1

Dimensions of Electric Potential per unit distance:

$V= \dfrac{\bigg[work\bigg]}{\bigg[charge\bigg]}$

$\implies V= \dfrac{\bigg[M{L}^{2} {T }^{ - 2} \bigg]}{\bigg[charge\bigg]}$

$\implies V = \dfrac{\bigg[M{L}^{2} {T }^{ - 3} \bigg]}{\bigg[ \dfrac{charge}{T}\bigg]}$

$\implies V= \dfrac{\bigg[M{L}^{2} {T }^{ - 3} \bigg]}{\bigg[ current\bigg]}$

$\implies V = \dfrac{\bigg[M{L}^{2} {T }^{ - 3} \bigg]}{\bigg[A\bigg]}$

$\implies V= \bigg[M{L}^{2} {T}^{ - 3} {A}^{ - 1} \bigg]$

Now, Potential per unit distance:

$\implies V/L=\dfrac{ \bigg[M{L}^{2} {T}^{ - 3} {A}^{ - 1} \bigg]}{L}$

$\implies V/L= \bigg[M{L}^{1} {T}^{ - 3} {A}^{ - 1} \bigg]$

Hope It Helps.