Saved Bookmarks
| 1. |
Four point charges Q , q Q and q are placed at the corners of a square of side 'a' as shown in the Fig. Find the resultant electric force on a charge Q |
|
Answer» Solution :Let us calculate electric force on a charge Q situated at the corner A in figure , then `|vecF_(AB)| = (1)/(4 pi in_(0)) (Qq)/(q^(2)) , |vecF_(AC)| = (1)/(4pi in_(0)) * (Q^(2))/(2A^(2)) ` and `|vecF_(AD)| = (1)/(4pi in_(0)) * (Qq)/(a^(2))` Directions of the force are as shown in Fig. `therefore` Force along BA , `F_(1) = F_(AB) + F_(AC) sin 45^(@)` `= (1)/(4pi in_(0)) (Qq)/(a^(2)) + (1)/(4 pi in_(0)) * (Q^(2))/(2a^(2)) (1)/(sqrt2)` `= (Q)/(4pi in_(0) a^(2)) [ q + (Q)/(2sqrt2)]` and force along `DA, F_(2) = F_(AD) + F_(AC) cos 45^(@)` =` (1)/(4 pi in_(0)) (Qq)/(a^(2)) + (1)/(4pi in_(0)) * (Q^(2))/(2a^(2)) (1)/(sqrt2)` `= (Q)/(4 pi in_(0) a^(2)) [ a + (Q)/(2 sqrt2)]` `therefore` Resultant electric force `F_(A)= sqrt(F_(1)^(2) - F_(2)^(2)) = (Q)/(4 pi in_(0) a^(2)) [ q + (Q)/(2sqrt2)] sqrt2 = (Q)/(8 pi in_(0) a^(2)) [2 sqrt2q + Q]` The resultant force acts along the EXTENDED line CA.
|
|