four charges +2q,-2q,-3q and +3q are kept in the corners of a square of side a.the total field at centre o is,
Answers
Answer:
Explanation:
Answer:
Explanation:
=> Electric field at point P due to charge +2q at A:
E_A = k(2q)/(a^2/2) along AP
=> Electric field at point P due to charge -2q at B:
E_B = k(-2q)/(a^2/2) along BP
=> Electric field at point P due to charge -3q at C:
E_C = k(-3q)/(a^2/2) along CP
=> Electric field at point P due to charge +3q at D:
E_D = k(3q)/(a^2/2) along DP
=> Electric field E_A and E_C are in opposite direction, thus, the net field is:
E_CA = k(-3q)/(a^2/2) -k(2q)/(a^2/2) = k(-5q)/(a^2/2) along CA
=> Electric field E_D and E_B are in opposite direction, thus, the net field is:
E_DB = k(3q)/(a^2/2) - k(-2q)/(a^2/2) = k(5q)(a^2/2) along DB
=> As, E_CA and E_DB are same in magnitude and perpendicular to each other, thus, the resultant E will bisect angle APB.
Thus, E = [(E_CA)^2 + (E_DB)^2]^1/2
E = [{k(-5q)/(a^2/2)}^2 +{k(5q)(a^2/2)}]^1/2
∴ E = √2k/a^2
