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Question:

Two point charges $q_A = 3 \mu C$ and $q_B = 3 \mu C$ are located 20 cm apart in vacuum.


(a) What is the electric field at the midpoint O of the line AB joining the two charges?

(b) If a negative test charge of magnitude $1.5 \times 10^{-9}C$ is placed at this point, what is the force experienced by the test charge?

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Last updated date: 13th Jun 2024
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Answer
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Hint:

(a) To find the electric field at point O, consider contributions from both charges, $q_A$ and $q_B$, using Coulomb's law. Take into account the direction of the electric fields.


(b) To calculate the force on a test charge at O, use the formula $\vec{F} = q_t \cdot \vec{E}$, where $q_t$ is the magnitude of the test charge, and $\vec{E}$ is the electric field at that location.


Step-by-Step Solution:

(a) Electric Field at the Midpoint O:

The situation is represented in the given figure, where O is the midpoint of line AB.

Distance between the two charges, $ AB = 20$cm.

Therefore, $AO = OB = 10$cm.

The net electric field at point O is denoted as $\vec{E}$.

Electric field at point O caused by the  $+3\mu C$  charge  $\vec{E_1}$ :
$\vec{E_1} = \dfrac{1}{4\pi\epsilon_0} \cdot \dfrac{q_A}{r_A^2} \hat{r_A}$

Where:

$q_A = 3\mu C$  (charge at point A)

$r_A = 10$ cm (distance from point A to  O)

The magnitude of the electric field at point O caused by the $ +3\mu C $ charge is:

$\vec{E_1} = \dfrac{1}{4\pi\epsilon_0} \cdot \dfrac{3 \times 10^{-6} C}{(0.1 m)^2} \hat{r_A}$ 

Similarly, calculate the electric field $\vec{E_2}$ at point  O at point O caused by the $-3\mu C$ charge:

$\vec{E_2} = \dfrac{1}{4\pi\epsilon_0} \cdot \dfrac{q_B}{r_B^2} \hat{r_B}$ 

Where:

$q_B = -3\mu C $ (charge at point B)

$r_B = 10$ (distance from point B to O)

The magnitude of the electric field at point O caused by the $ -3\mu C $ charge is:

$\vec{E_2} = \dfrac{1}{4\pi\epsilon_0} \cdot \dfrac{-3 \times 10^{-6} C}{(0.1 m)^2} \hat{r_B}$

The total electric field $ \vec{E} $ at O is the vector sum of $ \vec{E_1} $ and $ \vec{E_2} $ because they have the same magnitude and direction:

$\vec{E} = \vec{E_1} + \vec{E_2}$

The value of $ \vec{E} $ is obtained by multiplying 2 with the magnitude of $ \vec{E_1} $ or $ \vec{E_2} $ because they are equal:

$\vec{E} = 2 \times \dfrac{1}{4\pi\epsilon_0} \cdot \dfrac{3 \times 10^{-6} C}{(0.1 m)^2} \hat{r_A}$

$\vec{E} = 5.4 \times 10^6 \text{N/C along OB}$


(b) Force Experienced by the Test Charge:

The value of $ \vec{E} $ is obtained by multiplying 2 with the magnitude of $ \vec{E_1} $ or $ \vec{E_2} $ because they are equal:

$\vec{E} = 2 \times \dfrac{1}{4\pi\epsilon_0} \cdot \dfrac{3 \times 10^{-6} C}{(0.1 m)^2} \hat{r_A}$

$\vec{E} = 5.4 \times 10^6 \text{N/C along OB}$


Note: The electric field at the midpoint O is directed away from the midpoint, because both charges are positive. The force on the negative test charge is also directed away from the midpoint, because the test charge is negative and the electric field is directed away from the midpoint.