Chapter 21, Problem 102A

To determine

the magnitude and the direction of the electric field strength at a point above the x axis.

Answer to Problem 102A

The magnitude of the electric field is $8.1\times 10^4\text{N/C}$ and the direction will be towards the negative charge since it has the larger magnitude.

Explanation of Solution

Given:

Charge of the sphere A is $3.00\times {10}^{-6}\text{C}$ .

Charge of the sphere B is $-5.00\times {10}^{-6}\text{C}$ .

Separation between the sphere A and B is 0.800 m.

The given figure is shown below:

  

Formula used:

The magnitude of the force on the charge $q^{\prime }$ caused by the charge $q$ is defined as

  $F=K\frac{q{q}^{\prime }}{r^2}$  ......(1)

Here, $r$ is the distance between the given two charges, and $k=9.0\times {10}^9\text{N}\cdot {\text{m}}^{\text{2}}{\text{/C}}^{\text{2}}$ .

Electric field strength can be expressed as

  $E=\frac{F}{q^{\prime }}$  ......(2)

Calculation:

The electric field strength at a point due to two charges can be calculated by taking the vector sum of the electric fields due to the charges.

Substitute the value of $F$ in equation (2)

  $E_A=K\frac{q{q}^{\prime }}{r^2{q}^{\prime }}$

  $E_A=K\frac{q}{r^2}$  ......(3)

Substitute the values of $K$ , $q$ and $r$ in equation (3)

  $\begin{array}{l}{E}_A=\left(9.0\times {10}^9\text{N}\cdot {\text{m}}^{\text{2}}{\text{/C}}^{\text{2}}\right)\frac{\left(3.00\times {10}^{-6}\text{C}\right)}{{\left(0.800\text{m}\right)}^2}\\ E_A=4.2\times {10}^4\text{N/C}\end{array}$

Since the charge A is positive, the direction of the electric field will be away from the charge.

Now, find the field strength $E_B$ due to charge B ,

Substitute the values of $K$ , charge of the sphere B and $r$ in equation (3)

  $\begin{array}{l}{E}_A=\left(9.0\times {10}^9\text{N}\cdot {\text{m}}^{\text{2}}{\text{/C}}^{\text{2}}\right)\frac{\left(-5.00\times {10}^{-6}\text{C}\right)}{{\left(0.800\text{m}\right)}^2}\\ E_A=-7.0\times {10}^4\text{N/C}\end{array}$

The negative sign indicates that the direction of the electric field is towards the charge B .

The electric field strength $E$ due to charge A and charge B is given by

  $E=\sqrt{E_A^2+E_B^2}$  ......(4)

The resultant electric field strength is calculated as

  $\begin{array}{l}E=\sqrt{{\left(4.2\times {10}^4\text{N/C}\right)}^2+\left(-7.0\times {10}^4\text{N/C}\right)}\\ \boxed{E=8.1\times {10}^4\text{N/C}}\end{array}$

The direction will be towards the negative charge since it has the larger magnitude.

Conclusion:

Hence, magnitude of the electric field is $8.1\times 10^4\text{N/C}$ and the direction will be towards the negative charge since it has the larger magnitude.