question_answer

 

If a point charge \[+\,q\] is taken first from A to C and then from C to B of a circle drawn with another point charge \[+\,q\] at the centre, in below figure, then along which path, more work will be done?

Or

A point charge causes an electric flux of \[-1\times {{10}^{3}}N{{m}^{2}}{{C}^{-1}}\]  to pass through a spherical Gaussian surface of 10 cm radius centered on the charge.

(i) How much flux will pass through the surface, if the radius of the Gaussian surface is doubled?

(ii) Find the value of the point charge.

 

Answer:

As, charge +q is at the centre of the circle, therefore \[{{V}_{A}}={{V}_{B}}.\] If \[{{V}_{C}}\] is potential of point C, then \[{{V}_{C}}-{{V}_{A}}={{V}_{C}}={{V}_{B}}\] Or (i) same             (ii) \[q=-\,8.8\,n\,C\] Since, the charge +q is at the centre of the circle, therefore, \[{{V}_{A}}-{{V}_{B}}\] If \[{{V}_{C}}\] is potential of point C, then             \[{{V}_{C}}-{{V}_{A}}={{V}_{C}}-{{V}_{B}}\] or         \[{{W}_{AC}}=q({{V}_{C}}-{{V}_{A}})\]       \[(\because W=qV)\] also       \[{{W}_{CB}}=q({{V}_{B}}-{{V}_{C}})\]             \[=-q({{V}_{C}}-{{V}_{B}})=-q({{V}_{C}}-{{V}_{A}})\] Or

(i) Same, since the charge enclosed in both cases is same, hence amount of flux does not change.

(ii) As, we know \[{{\phi }_{E}}=\frac{q}{{{\varepsilon }_{0}}}\](Gauss?s law)

\[q={{\phi }_{E}}{{\varepsilon }_{0}}=-1\times {{10}^{3}}\times 8.85\times {{10}^{-12}}\approx -\,8.8\,\,nC\]