A) less than that before the balls touched
B) greater than that before the balls touched
C) same as that before the balls touched
D) zero
Correct Answer: B
Solution :
Original charges on spheres A and B be \[{{q}_{1}}\] and \[{{q}_{2}}\] respectively. Distance between the two spheres \[=r\] Since, both the spheres are of same size, they will possess equal charges on being brought in contact. \[\therefore \] \[q{{'}_{1}}=\frac{{{q}_{1}}+{{q}_{2}}}{2}\] Similarly, \[q{{'}_{2}}=\frac{{{q}_{1}}+{{q}_{2}}}{2}\] Therefore, new force of repulsion between spheres A and B is \[F'=\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{\left[ \frac{{{q}_{1}}+{{q}_{2}}}{2} \right]\left[ \frac{{{q}_{1}}+{{q}_{2}}}{2} \right]}{{{r}^{2}}}\] \[=\frac{{{\left[ \frac{{{q}_{1}}+{{q}_{2}}}{2} \right]}^{2}}}{4\pi {{\varepsilon }_{0}}{{r}^{2}}}\] As \[{{\left[ \frac{{{q}_{1}}+{{q}_{2}}}{2} \right]}^{2}}>\,{{q}_{1}}{{q}_{2}}\] \[\therefore \] \[F'>F\]
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