Practice magnitude of electric field just outside the conductor, electric field due to the point charge inside the cavity. Learn how to calculate electric field due to a charged conductor.
Proof
Since, Electric field inside conductor is zero, i.e., \(\vec E=0\)
Thus, \(\int\vec E\cdot d\vec A=0\) ....(1)
By Gauss's Law,
\(\int\vec E\cdot d\vec A=\dfrac {q_{inside}}{\epsilon_0}\)
\(0=\dfrac {q_{inside}}{\epsilon_0}\) [ From equation (1) ]
\(q_{inside}=0\)
Thus, there is no charge at point P and hence no electric field.
A Zero
B \(\dfrac {1}{4\pi\epsilon_0}\,\cdot \dfrac {Q}{r^2}\)
C \(\dfrac {1}{4\pi\epsilon_0}\,\cdot \dfrac {Q}{R^2}\)
D None of these
\(\vec E=0\)
\(\Rightarrow \oint\vec E \cdot d\vec A=0\)
By Gauss's Law,
\(\oint\vec E \cdot d\vec A=\dfrac {q_{inside}}{\mathcal{\epsilon}_0}\)
\(0=\dfrac {q_{inside}}{\mathcal{\epsilon}_0}\)
\(q_{inside}=0 \)
Proof
Since, Electric field inside conductor is zero, i.e., \(\vec E=0\).
Thus, \(\int\vec E\cdot d\vec A=0\) ....(1)
By Gauss's Law,
\(\int\vec E\cdot d\vec A=\dfrac {q_{inside}}{\epsilon_0}\)
\(0=\dfrac {q_{inside}}{\epsilon_0}\) [ From equation (1) ]
\(q_{inside}=0\)
Thus, there is no charge at point P and hence no electric field.
A Charge will distribute on its surface uniformly
B Charge will distribute throughout the volume of conductor uniformly
C Charge will distribute throughout the volume of conductor non-uniformly
D Charge will collect at a point on surface
\(Flux=E.A\;cos\,0^\circ\)
\(=\,EA\)
By Gauss's Law,
\(\text{Flux}=\dfrac {\sum Q_{enclosed}}{\epsilon_0}\) ...(1)
where, \(\sigma\) is surface charge density of the Gaussian surface where electric field is to be calculated
Thus, \(\text {Flux}=\dfrac {\sigma A}{\epsilon_0}\) ...(2)
Comparing equation (1) and (2)
\(EA=\dfrac {\sigma A}{\epsilon_0}\)
\(E=\dfrac {\sigma }{\epsilon_0}\)
\(\vec E=\dfrac {\sigma}{\epsilon_0}\;(\hat n)\), where \(\hat n\) is the unit vector in the direction of electric field
A \(20\,N/C\)
B \(30\,N/C\)
C \(1\,N/C\)
D \(Zero\)