Name: Continuous Charge Distribution
Uploaded: 2019-07-19
Duration: 5 min 39 s
Description: This video will help us understand the necessity of continuous charge distribution and learn about the concept of linear charge density. Learn more.

Question

Two points charges $\pm q$ are placed on the axis at $x = - a$ and $x = + a$ , as shown in the figure.

(a) Plot the variation of $E$ along the $x$ -axis.

(b) Plot the variation of $E$ along the $y$ -axis.

Accepted Answer

Electric field due to a point charge is given as $E = \frac{k q}{x^{2}}$ , where $k$ is constant, $q$ is charge and $x$ is distance from the charge.

For $x = \infty$ and $x = - \infty$ , magnitude of electric field is $E = 0$ .

For $x = 0$ or very close to the charge, magnitude of electric field is $E = \infty$ .

For a positive point charge, the electric field due to it originates from it and goes away from it in a straight line and for negative point charge, the electric field moves towards it in a straight line.

For the positive charge, the electric field will be leftwards towards the left of charge and rightwards towards the right of charge. For the negative charge, the electric field will be rightwards towards the left of charge and leftwards towards the right of charge. At the origin, both the electric fields are in the same direction. Hence, the magnitude of electric field at the origin is given as $E_{0} = \frac{k q}{{(a)}^{2}} + \frac{k q}{{(a)}^{2}} = \frac{2 k q}{a^{2}}$ in the rightwards direction.

If we plot the electric field towards the positive $x$ -direction as positive and negative in the opposite direction then,

Since the magnitude of charge is same and the any point on $y$ -axis is equidistant from the charges. Hence, the magnitude of electric field will same for both charges. Let it be equal to $E .$

From the above diagram, the resultant electric field due to charges is given as $E_{r} = 2 E \cos θ$ , where $θ$ is angle made by resultant and electric field due to one charge.

From the diagram, we can see that, $- \frac{π}{2} < θ < \frac{π}{2}$ .

$\Rightarrow 0 < \cos θ < 1 \Rightarrow 0 < 2 E \cos θ < 2 E \Rightarrow 0 < E_{r} < 2 E$

Hence, the resultant electric field of the charges varies from $0$ at infinity to maximum of $2 E = 2 \frac{k q}{a^{2}}$ at the origin.

Since the direction of resultant electric field is in rightwards direction and as per our assumption, it will be positive. Hence, the variation of electric field in $y$ -direction is given as shown below.

Two points charges $\pm q$ are placed on the axis at $x = - a$ and $x = + a$ , as shown in the figure.

(a) Plot the variation of $E$ along the $x$ -axis.

(b) Plot the variation of $E$ along the $y$ -axis.

Important Questions on Coulomb's Law and Electric Field

Learn and Prepare for any exam you want !

Two points charges ±q are placed on the axis at x=-a and x=+a, as shown in the figure. (a) Plot the variation of E along the x-axis. (b) Plot the variation of E along the y-axis.

Important Questions on Coulomb's Law and Electric Field

Learn and Prepare for any exam you want !

Two points charges $\pm q$ are placed on the axis at $x = - a$ and $x = + a$ , as shown in the figure.

(a) Plot the variation of $E$ along the $x$ -axis.

(b) Plot the variation of $E$ along the $y$ -axis.