Saved Bookmarks
| 1. |
In the previous problem, if we connect the conductors 1 and 2, 2 and 3, 1 and 3, 1 and ground, 2 and ground, and 3 and ground, separately, find the charge flown in each case. |
|
Answer» Solution :(i) If we connect the conductors 1 and 2, the potential of spheres 1 and 2 will be equal. Here `V_1 = V_2` or `K (x)/a + K (Q - x)/(2 a) + K((-4Q))/(3 a)` =`K(x)/a + K(Q - x)/(2 a) + K((-4 Q))/(3 a)` or `x = 0` The CHARGE of sphere 1 will come to sphere 2. Hence charge flown in this case is equal to `-Q`.  . (ii) If we connect the spheres 2 and 3, their potential will be same. `V_2 = V_3` or `K((-Q))/(2 a) + K(x)/(2 a) + K((-2 Q - x))/(3 a)` =`K ((-Q))/(3 a) + K(x)/(3 a) + K((2 Q - x))/(3 a)` or `x = Q` Hence, charge of amount Q will FLOW from 2 to 3.  . (iii) If we 1 and 3, their potential should be equal. `V_1 = V_3` or `K(x)/a + K(2 Q)/(2 a) + K((-5 Q - x))/(3 a)` =`K(x)/(3 a) + K(2 Q)/(3 a) + K((-5 Q - x))/(3 a)` or `(x)/a - (x)/(3 a) = (2 Q)/(3 a) - (2 Q)/(2 a)` or `(2 x)/a = -Q/(3) or x = (-Q)/(2)` Hence, `-Q//2` charge will flow from 1 to 3.  . (iv) Now sphere 1 is connected with GROUND. The potential of sphere 1 will be zero. `V_1 = 0 = K (x)/a + K(2 Q)/(2 a) + K ((-4 Q))/(3 a)` or `x = -Q + (4 Q)/(3) = Q/(3)` Hence, `Q + Q/(3) = (4 Q)/(3)` should flow from ground to sphere.  . (v) Now sphere 2 is connected with earth. The potential sphere 2 should be zero. `V_2 = 0 = K((- Q))/(2 a) + K (x)/(2 a) + K((-4 Q))/(3 a)` or `(x)/(2) = Q/(2) + (4 Q)/(3)` or `x = Q + (8 Q)/(3) = (11 Q)/(3)` hence, `(11 Q)/(3) - 2 Q = (5 Q)/(3)` charge will flow from earth sphere.  . (vi) Now sphere 3 is connected with ground. Here, `V_3 = 0 = K ((-Q))/(3 a) + K ((2 Q))/(3 a) + K((x))/(3 a)` or `x = Q - 2 Q = -Q` Hence, `- 3 Q` charge will flow from sphere to earth or `3 Q` charge will flow from earth to sphere.  .
|
|