Question

Two point charges -Q and Q are located at points (d, 0) and (0, d) respectively, in x-y plane. The electric field E at the origin will be:

• A. \(\frac{1}{4\pi\epsilon_0} \frac{\sqrt{2}Q}{d^2} (\hat{i} - \hat{j})\)
• B. \(\frac{1}{4\pi\epsilon_0} \frac{\sqrt{2}Q}{d^2} (-\hat{i} - \hat{j})\)
• C. \(\frac{1}{4\pi\epsilon_0} \frac{Q}{d^2} (\hat{i} - \hat{j})\)
• D. \(\frac{1}{4\pi\epsilon_0} \frac{Q}{d^2} (-\hat{i} - \hat{j})\)

Hint:

Always draw the vectors first! For the origin, a field from a positive charge on the y-axis points down (\(-\hat{j}\)), and a field from a negative charge on the x-axis points right (\(+\hat{i}\)).

Answer 1

The Correct Option is A

To find the electric field \( \mathbf{E} \) at the origin due to two point charges -Q and Q located at points (d, 0) and (0, d) in the x-y plane, we follow these steps:

1. Understand the contribution of each charge to the electric field at the origin:
   • Electric field due to a point charge \( Q \) is given by the formula: \(E = \frac{1}{4\pi\epsilon_0} \frac{Q}{r^2} \hat{r}\) where \( r \) is the distance from the charge to the point and \( \hat{r} \) is the unit vector from the charge to the point.
2. Calculate the electric field at the origin due to charge -Q located at (d, 0):
   • Distance from charge to the origin: \( d \)
   • Unit vector \( \hat{r} \) from charge to origin: \(-\hat{i}\)
   • Electric field: \(E_1 = \frac{1}{4\pi\epsilon_0} \frac{Q}{d^2} \hat{i}\)
3. Calculate the electric field at the origin due to charge Q located at (0, d):
   • Distance from charge to the origin: \( d \)
   • Unit vector \( \hat{r} \) from charge to origin: \(-\hat{j}\)
   • Electric field: \(E_2 = \frac{1}{4\pi\epsilon_0} \frac{Q}{d^2} (-\hat{j})\)
4. Combine the electric fields from both charges:
   • Total electric field at origin: \(\mathbf{E} = E_1 + E_2 = \frac{1}{4\pi\epsilon_0} \left( \frac{Q}{d^2} \hat{i} + \frac{Q}{d^2} (-\hat{j}) \right) = \frac{1}{4\pi\epsilon_0} \frac{Q}{d^2} (\hat{i} - \hat{j})\)
5. Consider symmetry and alternative approaches to validate your solution:
   • The two charges are symmetric about the line \( y = x \), leading to field magnitude symmetry.
   • Therefore, the combined contribution results in equal components along \(\hat{i} - \hat{j}\).

Therefore, the electric field at the origin is:

\(\frac{1}{4\pi\epsilon_0} \frac{\sqrt{2}Q}{d^2} (\hat{i} - \hat{j})\)