Question

In the region where electric field E = 0, the relation between electirc potential V and distance r is

Answer 1

Answer:

Given:

Electric Field Intensity is zero in a particular region wrt to a source.

To find:

Relationship between Electric Potential V and distance r

Concept:

Electric Field intensity is a vector index representing the strength of the field produced by the source.

Whereas Electric Potential is a scalar index also representing the strength.

Calculation:

$\displaystyle{ \int \: dV = \int \: E \: dr}$

Putting the limits :

$\displaystyle{ \int_{V1}^{V2} dV = \int_{r1}^{r2} \: E \: dr}$

$\implies\displaystyle{ \int_{V1}^{V2} dV = \int_{r1}^{r2} \: 0 \times \: dr}$

$\implies V2 - V1 = 0$

$\implies V2 = V1$

$\implies V = constant$

So Electric Potential becomes constant .

It becomes independent of distance from the source.

$\: \: \: \: \: \: \: \: \: \: \: \: \boxed{ \huge{ \green{ \sf{ \bold{V \propto \: {r}^{0}}}}}}$

Answer 2

$\huge \underline {\underline{ \mathfrak{ \green{Ans}wer \colon}}}$

As we know that :

$\large{\boxed{\sf{dV \: = \: E \: dr}}}$

Ingrate Both the sides and put limits

$\implies {\sf{ \int dV \: = \: \int E dr}} \\ \\ \implies \displaystyle {\sf{ \int _{v_1} ^{v_2} dV \: = \: E \int _{r_1} ^{r_2} dr}} \\ \\ \implies {\sf{V_2 \: - \: V_1 \: = \: E \big[ r_2 \: - \: r_1 \big]}} \\ \\ \implies {\sf{V_2 \: - \: V_1 \: = \: 0(r_2 \: - \: r_1)}} \\ \\ \implies{\sf{V_2 \: - \: V_1 \: = \: 0}} \\ \\ \implies {\sf{V_2 \: = \: V_1}} \\ \\ \implies {\sf{V \: = \: constant}}$

So, Electric Potential became a constant

$\Large{\boxed{\boxed{\sf{V \: \propto \: r}}}}$