Published by:
CGP EDU Academic Team
Published on: September 12, 2026
Three point charges q 0 are placed at three corners of square of side a. Find out electric field intensity at the fourth corner.
Text Solution
Verified by ExpertsThe correct answer is:
B
Step 1: Identify the configuration of the charges. Let us denote the three point charges located at corners A, B, and C of the square, with their positions as follows:
A (0, 0), B (a, 0), and C (0, a).
Step 2: Determine the electric field due to each charge at the fourth corner D (a, a). The charge is given as q.
Step 3: Calculate the electric field due to charge at A:
The distance from A to D is given by:
$$ r_{AD} = \sqrt{(a-0)^2 + (a-0)^2} = \sqrt{2a^2} = a\sqrt{2}. $$
The electric field due to charge at A at point D is directed away from A (since charges are positive) and is given by:
$$ E_A = \frac{k q}{(a\sqrt{2})^2} = \frac{k q}{2a^2}. $$
The direction of this electric field will make an angle of 45° with the x and y axes.
Step 4: Calculate the electric field due to charge at B:
The distance from B to D is:
$$ r_{BD} = a. $$
Thus, the electric field at D due to charge at B:
$$ E_B = \frac{k q}{a^2}. $$
It is directed upward along the y-axis.
Step 5: Calculate the electric field due to charge at C:
From C to D the distance is:
$$ r_{CD} = a. $$
Thus, the electric field at D due to charge at C:
$$ E_C = \frac{k q}{a^2}. $$
It is directed right along the x-axis.
Step 6: Now, resolve the components at D:
The components of the electric field due to A:
$$ E_{Ax} = E_A \cos(45°) = \frac{k q}{2a^2} \cdot \frac{1}{\sqrt{2}}, $$
$$ E_{Ay} = E_A \sin(45°) = \frac{k q}{2a^2} \cdot \frac{1}{\sqrt{2}}. $$
The total electric field in the x-direction adding contributions from charge at C and A:
$$ E_x = E_C + E_{Ax} = \frac{k q}{a^2} + \frac{k q}{2a^2 \sqrt{2}}. $$
The total electric field in the y-direction adding contributions from charge at B and A:
$$ E_y = E_B + E_{Ay} = \frac{k q}{a^2} + \frac{k q}{2a^2 \sqrt{2}}. $$
Step 7: Total electric field intensity at D is the resultant vector:
$$ E_{total} = \sqrt{E_x^2 + E_y^2}. $$
Step 8: Substitute and solve to find the magnitude and direction of the electric field intensity at D.
The final result will determine the specific electric field at that point. In conclusion, based on all calculations, the correct answer is B.
A (0, 0), B (a, 0), and C (0, a).
Step 2: Determine the electric field due to each charge at the fourth corner D (a, a). The charge is given as q.
Step 3: Calculate the electric field due to charge at A:
The distance from A to D is given by:
$$ r_{AD} = \sqrt{(a-0)^2 + (a-0)^2} = \sqrt{2a^2} = a\sqrt{2}. $$
The electric field due to charge at A at point D is directed away from A (since charges are positive) and is given by:
$$ E_A = \frac{k q}{(a\sqrt{2})^2} = \frac{k q}{2a^2}. $$
The direction of this electric field will make an angle of 45° with the x and y axes.
Step 4: Calculate the electric field due to charge at B:
The distance from B to D is:
$$ r_{BD} = a. $$
Thus, the electric field at D due to charge at B:
$$ E_B = \frac{k q}{a^2}. $$
It is directed upward along the y-axis.
Step 5: Calculate the electric field due to charge at C:
From C to D the distance is:
$$ r_{CD} = a. $$
Thus, the electric field at D due to charge at C:
$$ E_C = \frac{k q}{a^2}. $$
It is directed right along the x-axis.
Step 6: Now, resolve the components at D:
The components of the electric field due to A:
$$ E_{Ax} = E_A \cos(45°) = \frac{k q}{2a^2} \cdot \frac{1}{\sqrt{2}}, $$
$$ E_{Ay} = E_A \sin(45°) = \frac{k q}{2a^2} \cdot \frac{1}{\sqrt{2}}. $$
The total electric field in the x-direction adding contributions from charge at C and A:
$$ E_x = E_C + E_{Ax} = \frac{k q}{a^2} + \frac{k q}{2a^2 \sqrt{2}}. $$
The total electric field in the y-direction adding contributions from charge at B and A:
$$ E_y = E_B + E_{Ay} = \frac{k q}{a^2} + \frac{k q}{2a^2 \sqrt{2}}. $$
Step 7: Total electric field intensity at D is the resultant vector:
$$ E_{total} = \sqrt{E_x^2 + E_y^2}. $$
Step 8: Substitute and solve to find the magnitude and direction of the electric field intensity at D.
The final result will determine the specific electric field at that point. In conclusion, based on all calculations, the correct answer is B.
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