Physics, asked by Vishal101100, 8 months ago

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Find the relationship b/w Electric field and potential difference!!!!!

Answers

Answered by nirman95

To derive :

Relationship between Electric Field and Electric Potential .

Proof:

First of all , let's consider a charge q located at a specified point and emitting electric field . We shall also consider 2 closely located points named as A and B.

The distance between A and B be dx.

Since the dx distance is extremely small, we can consider that the Electric field E (from the charge) remains same at both the points.

Now the work done to move a unit positive charge from A to B will be :

$\boxed{ \bold{dW = - E \times dx }}$

The negative sign denoted that the work done was against the Electric Field Direction.

For a unit positive charge , we know that work done us equal to the potential difference dV.

$\bigstar \: \: \bold{dV = - E \times dx }$

$\bold{ = > E = - \dfrac{dV}{dx} }$

So we can say that Electric Field is the gradient of Electric Potential with respect to distance.

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Answered by Mastershivansh007

Answer:

The relationship between potential and field (E) is a differential: electric field is the gradient of potential (V) in the x direction. This can be represented as: Ex=−dVdx E x = − dV dx . Thus, as the test charge is moved in the x direction, the rate of the its change in potential is the value of the electric field.

Explanation:

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Find the relationship b/w Electric field and potential difference!!!!!​

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To derive :

Proof:

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Find the relationship b/w Electric field and potential difference!!!!!