Point charges +q, -q, -q, q, +Q and -q are at the vertices of a hexagon. The E-field at O due to A, B, C, D, F is twice the field due to +Q at E. The value of Q is:
Show Hint
Symmetry in a hexagon usually results in the cancellation of fields from opposite identical charges.
Step 1: Understanding the Question:
We analyze the symmetry of a regular hexagon to find the net electric field at the center. Step 2: Key Formula or Approach:
Electric field \( E = \frac{kq}{r^2} \). In a regular hexagon, the distance from center to each vertex is the same (\( r \)). Step 3: Detailed Explanation:
Let \( E_0 = \frac{kq}{r^2} \) be the field magnitude due to a single charge \( q \).
By symmetry, a set of six equal charges at the vertices produces zero field at the center.
If one charge \( +q \) is removed from E, the field due to the remaining five is \( E_0 \) directed towards E.
Here, charges at opposite vertices A(\( +q \)) and D(\( +q \)) cancel each other.
Charges at B(\( -q \)) and F(\( -q \)) produce fields towards B and F. Resultant is \( E_0 \) towards E bisector.
Actually, the resultant of the 5 charges acts like a single charge \( -q \) at E, so field magnitude is \( E_{5} = \frac{kq}{r^2} \).
Field due to \( +Q \) at E alone is \( E_Q = \frac{kQ}{r^2} \).
Given \( E_{5} = 2 E_Q \).
\[ \frac{kq}{r^2} = 2 \left( \frac{kQ}{r^2} \right) \]
\[ q = 2Q \implies Q = \frac{q}{2} \] Step 4: Final Answer:
The value of \( Q \) is \( \frac{q}{2} \).
