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    				| 1. | As per this diagram a point charge `+q` is placed at the origin `O`. Work done in taking another point charge `-Q` from the point `A(0, a)` to another point `B(a, 0)` along the staight path `AB` is: A. `(qQ)/(4piepsi_(0))((a-b)/(ab))`B. `(qQ)/(4piepsi_(0))((b-a)/(ab))`C. `(qQ)/(4piepsi_(0))((b)/a^(2)-(1)/(b))`D. `(qQ)/(4piepsi_(0))((a)/(b^(2))-(1)/(b))` | 
| Answer» Correct Answer - A Potential at point A is `V_A=1/(4piepsilon_0) q/a` Potential at point B is `V_B=1/(4piepsilon_0) q/b` Work done in taking a charge Q from A to B is `W=Q(V_B-V_A)=(Qq)/(4piepsilon_0)[1/b-1/a]=(Qq)/(4piepsilon_0)["a-b"/"ab"]` | |