Physics : Electrostatics: Electrostatic Potential and Potential Energy : Relation between electric field and potential

**Relation between**** ****electric field and
potential**

Consider a positive
charge *q* kept fixed at the origin. To move a unit positive charge by a
small distance *dx* in the electric field E, the work done is given
by *dW* = −*E dx*. The minus sign implies that work is done against
the electric field. This work done is equal to electric potential difference.
Therefore,

The electric field is
the negative gradient of the electric potential. In general,

**EXAMPLE 1.14**

The following figure
represents the electric potential as a function of x – coordinate. Plot the
corresponding electric field as a function of x.

**Solution**

In the given problem, since the potential depends only on x, we can use (the other two terms ∂V/∂y and ∂V/∂z are zero)

From 0 to 1 cm, the
slope is constant and so dV/dx = 25V cm^{−1}.

So = −25V cm−1

From 1 to 4 cm, the
potential is constant, V = 25 V. It implies that dV/dx = 0. So = 0

From 4 to 5 cm, the
slope dV/d*x* = −25V cm^{−1}.

So = +25V cm−1

The plot of electric
field for the various points along the x axis is given below.

Tags : Electrostatics , 12th Physics : Electrostatics

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12th Physics : Electrostatics : Relation between electric field and potential | Electrostatics

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