Four point charges (with equal magnitude of charge of \( 5 \text{ C} \); but with different signs) are placed at four corners of a square of side \( 10 \text{ m} \). Assuming the square is centered at the origin, the potential and the magnitude of electric field at the origin are:
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Always check for symmetry[ 3]. If charges are equal and opposite across a point, the potential will be zero, but the electric field vectors might reinforce each other.
$0 \text{ V}$ and $\frac{\sqrt{2}}{5}k \text{ V/m}$
$\frac{\sqrt{2}}{5}k \text{ V}$ and $\frac{\sqrt{2}}{5}k \text{ V/m}$
$0 \text{ V}$ and $5 \text{ V/m}$
$\frac{\sqrt{2}}{5}k \text{ V}$ and $0 \text{ V/m}$
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The Correct Option isB
Solution and Explanation
Concept:
Electric potential is a scalar sum, while the electric field is a vector sum[ 14].
$$V = \sum \frac{kq_i}{r_i}, \quad \vec{E} = \sum \vec{E_i}$$
Step 1: {Determine the distance $r$ from the corners to the center.}
Side $s = 10 \text{ m}$. Half-diagonal $r = \frac{10\sqrt{2}}{2} = 5\sqrt{2} \text{ m}$.
Step 2: {Calculate net potential $V$ at the origin.}
$V = \frac{k}{r}(+5 - 5 + 5 - 5) = 0 \text{ V}$.
Step 3: {Calculate the magnitude of the net electric field $E$.}
Each charge produces a field $E_c = \frac{k(5)}{(5\sqrt{2})^2} = \frac{5k}{50} = \frac{k}{10}$. Based on the configuration, the vectors add up to:
$$E_{net} = \sqrt{(2E_c)^2 + (2E_c)^2} = 2E_c\sqrt{2} = 2\left(\frac{k}{10}\right)\sqrt{2} = \frac{\sqrt{2}}{5}k \text{ V/m}$$
