
In the direction of electric field, the electric potential:
A. Decreases.
B. Increases.
C. Remains unchanged.
D. Becomes zero.
Answer
597.3k+ views
Hint: We will use the idea of the electric field as a gradient of potential to answer this problem. We are going to solve this problem using the relationship \[E = - \dfrac{{dv}}{{dr}}\].
Complete step by step answer:
We know that in an electrical field the electrical potential between two points A and B is defined as the amount of work performed per unit of positive test charge in moving it against the electrostatic force due to the electrical field from point A to point B.
If $dW$ is the work done in moving a small positive test charge ${q_0}$ from point P to Q, then
$ \Rightarrow (V + dV) - V = \dfrac{{dW}}{{{q_0}}}$.
$ \Rightarrow dV = \dfrac{{dW}}{{{q_0}}}$ ………..(i)
And, if $\overrightarrow E $ is the electric field at point P due to charge +a place at point O, then the test charge experiences a force equal to ,
$ \Rightarrow \overrightarrow F = {q_0}\overrightarrow E $.
Therefore, work done to move the test charge through an infinitely small displacement $PQ = dl$is given by
$dW = \overrightarrow F .\overrightarrow {dl} $
and we have $\overrightarrow F = {q_0}\overrightarrow E $, then
$dW = {q_0}\overrightarrow E .\overrightarrow {dl} = ({q_0}\overrightarrow E ).\overrightarrow {dl} \cos {180^0}$ [ ]
$dW = - {q_0}E.dl$ ……..(ii)
As the distance r decreases in the direction of $\overrightarrow {dl} $, the distance dl is taken as dr.
So now, form equation (ii), we get
$
dW = - {q_0}E.dr \\
\dfrac{{dW}}{{{q_0}}} = - E.dr \\
$
From equation (i), we have $dV = \dfrac{{dW}}{{{q_0}}}$, then
$
dV = - Edr \\
E = - \dfrac{{dV}}{{dr}} \\
$.
Therefore the electric field at a point is equal to the negative gradient of the electric potential at that point.
The negative sign indicates that the direction of E is always in the direction of decreased potential.
Hence, we can say that, In the direction of electric field, the electric potential decreases.
Therefore the correct answer is option (A).
Note: In this type of questions, first we have to identify what relationship of the given terms is required to solve the question. Then we will assume some conditions and prove the statement step by step. While solving the question, we have to remember the significance of the signs. After that we will get the required answer.
Complete step by step answer:
We know that in an electrical field the electrical potential between two points A and B is defined as the amount of work performed per unit of positive test charge in moving it against the electrostatic force due to the electrical field from point A to point B.
If $dW$ is the work done in moving a small positive test charge ${q_0}$ from point P to Q, then
$ \Rightarrow (V + dV) - V = \dfrac{{dW}}{{{q_0}}}$.
$ \Rightarrow dV = \dfrac{{dW}}{{{q_0}}}$ ………..(i)
And, if $\overrightarrow E $ is the electric field at point P due to charge +a place at point O, then the test charge experiences a force equal to ,
$ \Rightarrow \overrightarrow F = {q_0}\overrightarrow E $.
Therefore, work done to move the test charge through an infinitely small displacement $PQ = dl$is given by
$dW = \overrightarrow F .\overrightarrow {dl} $
and we have $\overrightarrow F = {q_0}\overrightarrow E $, then
$dW = {q_0}\overrightarrow E .\overrightarrow {dl} = ({q_0}\overrightarrow E ).\overrightarrow {dl} \cos {180^0}$ [ ]
$dW = - {q_0}E.dl$ ……..(ii)
As the distance r decreases in the direction of $\overrightarrow {dl} $, the distance dl is taken as dr.
So now, form equation (ii), we get
$
dW = - {q_0}E.dr \\
\dfrac{{dW}}{{{q_0}}} = - E.dr \\
$
From equation (i), we have $dV = \dfrac{{dW}}{{{q_0}}}$, then
$
dV = - Edr \\
E = - \dfrac{{dV}}{{dr}} \\
$.
Therefore the electric field at a point is equal to the negative gradient of the electric potential at that point.
The negative sign indicates that the direction of E is always in the direction of decreased potential.
Hence, we can say that, In the direction of electric field, the electric potential decreases.
Therefore the correct answer is option (A).
Note: In this type of questions, first we have to identify what relationship of the given terms is required to solve the question. Then we will assume some conditions and prove the statement step by step. While solving the question, we have to remember the significance of the signs. After that we will get the required answer.
Recently Updated Pages
Two men on either side of the cliff 90m height observe class 10 maths CBSE

What happens to glucose which enters nephron along class 10 biology CBSE

Cutting of the Chinese melon means A The business and class 10 social science CBSE

Write a dialogue with at least ten utterances between class 10 english CBSE

Show an aquatic food chain using the following organisms class 10 biology CBSE

A circle is inscribed in an equilateral triangle and class 10 maths CBSE

Trending doubts
The shortest day of the year in India

Why is there a time difference of about 5 hours between class 10 social science CBSE

Write a letter to the principal requesting him to grant class 10 english CBSE

What is the median of the first 10 natural numbers class 10 maths CBSE

The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths

What is the missing number in the sequence 259142027 class 10 maths CBSE

