Problem 103 - A Complete Resource Book in Physics for JEE Main

Question

Charges \(Q_{1}\) and \(Q_{2}\) lie inside and outside, respectively, of a closed Gaussian surface \(S .\) Let \(E\) be the field at any point on \(S\) and \(\phi\) be the flux of \(E\) over \(S\), (A) If \(Q_{1}\) changes, both \(E\) and \(\phi\) will change. (B) If \(Q_{2}\) changes, \(E\) will change. (C) If \(Q_{1}=0\) and \(Q_{2} \neq 0\) then \(E \neq 0\) but \(\phi=0\). (D) If \(Q_{1} \neq 0\) and \(Q_{2}=0\) then \(E=0\) but \(\phi \neq 0\).

Step-by-Step Solution

Answer

In conclusion, out of the given statements: - Statement A is correct. - Statement B is incorrect. - Statement C is partially correct. - Statement D is partially correct.

Step 1: Statement A:

If \(Q_1\) changes, both \(E\) and \(\phi\) will change. To verify this statement, we need to recall Gauss's law: \[\phi = \oint_{S} \vec{E} \cdot \vec{dA} = \frac{Q_{\text{enclosed}}}{\varepsilon_0}\] Given that \(Q_1\) lies inside the Gaussian surface, changing the value of \(Q_1\) will change the total charge enclosed by the surface. The electric field E at any point on S is dependent on the enclosed charge. Therefore, if \(Q_1\) changes, E will change too. As \(\phi\) is directly proportional to the enclosed charge, changing \(Q_1\) will also cause a change in the electric flux \(\phi\). Thus, statement A is correct.

Step 2: Statement B:

If \(Q_2\) changes, \(E\) will change. In this case, we must remember that the electric field E at any point on the Gaussian surface S is only dependent on the enclosed charge. Since \(Q_2\) lies outside the Gaussian surface, changes in its value do not affect the enclosed charge and thus won't affect the electric field E at any point on the surface S. Therefore, statement B is incorrect.

Step 3: Statement C:

If \(Q_1 = 0\) and \(Q_2 \neq 0\) then \(E \neq 0\) but \(\phi = 0\). Here, we need to analyze two conditions: 1. \(Q_1 = 0\) and \(Q_2 \neq 0\) 2. \(E \neq 0\) but \(\phi = 0\) If \(Q_1 = 0\), the enclosed charge becomes zero. However, \(Q_2\) is not necessarily equal to zero. We must remember that the field E at the surface S is affected only by the enclosed charge. As there is no enclosed charge due to \(Q_1 = 0\), the electric field E will be equal to zero, contrary to what Statement C says. Since the enclosed charge is zero, Gauss's law gives us \(\phi = 0\), which is in line with Statement C. Thus, statement C is partially correct (as only the last part, \(\phi = 0\), is true).

Step 4: Statement D:

If \(Q_1\neq 0\) and \(Q_2 = 0\) then \(E = 0\) but \(\phi \neq 0\). In this case, we again need to analyze two conditions: 1. \(Q_1 \neq 0\) and \(Q_2 = 0\) 2. \(E = 0\) but \(\phi \neq 0\) If \(Q_1 \neq 0\), this charge is enclosed by the Gaussian surface, and therefore, it will affect the electric field E. So, E will not be equal to 0, contrary to what Statement D says. Now, considering Gauss's law, since \(Q_1 \neq 0\), the electric flux \(\phi\) will not be equal to 0, which is in line with Statement D. Thus, statement D is partially correct (as only the last part, \(\phi \neq 0\), is true).