Five positive charges each having charge q are placed at the vertices of a pentagon as shown in the figure. The electric potential (V) and the electric field ( E) at the center O of the pentagon due to these five positive charges are : [figure]
- V = 5q4 0 r and E = 5 3q8 0 r2 r
- V = 5q4 0 r and E = 0
- V = 0 and E = 0
- V = 5q4 0 r and E = 5q4 0 r2 r
Answer
(2) V = 5q4 0 r and E = 0
Solution
Each charge is at the same distance r from the centre O (regular pentagon).
Potential is a scalar, so the five contributions add:
V = 5 q4 0 r = 5q4 0 r
Field is a vector: the five fields at O have equal magnitude q4 0 r2 and are directed along five directions 72 apart.
By symmetry (rotating the figure by 72 leaves it unchanged), their resultant must be zero:
E = 0
Answer: (2)