Question 834: Electrostatics & Capacitance

Question

A point charge $$+Q$$ is placed just outside an imaginary hemispherical surface of radius $$R$$ as shown in the figure.  Which of the following statements is/are correct?

Accepted Answer

Since charge Q is outside the hemispherical surface, the net flux passing through the curved surface of hemispherical surface and flat surface is zero.
Therefore,
$$\phi$$
curved
+ $$\phi$$
flat
= 0 ....... (1)
Hence, option (B) is incorrect.
Now,
$${\phi _{flat}} = \int {\overrightarrow E .\,d\overrightarrow A  = \int {EdA\cos 	heta } } $$
and $$E = {1 \over {4\pi {\varepsilon _0}}}{Q \over {{{(\sqrt {{R^2} + {r^2}} )}^2}}} = {1 \over {4\pi {\varepsilon _0}}}{Q \over {({R^2} + {r^2})}}$$
Also,
$$\cos 	heta  = {R \over {\sqrt {{R^2} + {r^2}} }}$$
and
$$A = 2\pi r \Rightarrow dA = 2\pi rdr$$
Therefore,
$${\phi _{flat}} = \int\limits_0^R {{1 \over {4\pi {\varepsilon _0}}}{Q \over {{R^2} + {r^2}}}2\pi rdr{R \over {\sqrt {{R^2} + {r^2}} }}} $$
$${\phi _{flat}} = {{QR} \over {2{\varepsilon _0}}}\int\limits_0^R {{{rdr} \over {{{({R^2} + {r^2})}^{3/2}}}}} $$
Substituting R
2
+ r
2
= t, we get
$$2rdr = dt$$
$$ \Rightarrow {\phi _{flat}} = {{QR} \over {2{\varepsilon _0}}}\int\limits_0^R {{{dt} \over 2}{1 \over {{t^{3/2}}}} = {{QR} \over {2{\varepsilon _0}}}\left[ {{1 \over 2}{{{t^{ - 1/2}}} \over { - 1/2}}} ight]_0^R} $$
Substituting $$t = {R^2} + {r^2}$$, we get
$${\phi _{flat}} = {{QR} \over {2{\varepsilon _0}}}\left[ {{1 \over 2}{{{{({R^2} + {r^2})}^{ - 1/2}}} \over { - 1/2}}} ight]_0^R$$
$$ = {{QR} \over {2{\varepsilon _0}}}\left[ {{{ - 1} \over {\sqrt {{R^2} + {r^2}} }}} ight]_0^R = {{QR} \over {2{\varepsilon _0}}}\left( {{{ - 1} \over {\sqrt {{R^2} + {R^2}} }} + {1 \over {\sqrt {{R^2}} }}} ight)$$
$$ = {{QR} \over {2{\varepsilon _0}}}\left( {{{ - 1} \over {\sqrt {2{R^2}} }} + {1 \over {\sqrt {{R^2}} }}} ight) = {{QR} \over {2{\varepsilon _0}}}\left( {{{ - 1} \over {\sqrt 2 R}} + {1 \over R}} ight)$$
$$ = {{QR} \over {2{\varepsilon _0}}}{1 \over R}\left( {{{ - 1} \over {\sqrt 2 }} + 1} ight) = {Q \over {2{\varepsilon _0}}}\left( {{{ - 1} \over {\sqrt 2 }} + 1} ight)$$
Using Eq. (1), we get
$${\phi _{curved}} =  - {\phi _{flat}} =  - {Q \over {2{\varepsilon _0}}}\left( {{{ - 1} \over {\sqrt 2 }} + 1} ight) =  - {Q \over {2{\varepsilon _0}}}\left( {1 - {1 \over {\sqrt 2 }}} ight)$$
Hence, option (A) is correct.
The potential at any point on the circumference of the flat surface is $${1 \over {4\pi {\varepsilon _0}}}{Q \over {\sqrt 2 R}}$$.
Thus, the circumference of flat surface is equipotential.

Answer

The electric flux passing through the curved surface of the hemisphere is
$$ - {Q \over {2{\varepsilon _0}}}$$ $$\left( {1 - {1 \over {\sqrt 2 }}} \right)$$

Answer

Total flux through the curved and the flat surfaces is $${Q \over {{\varepsilon _0}}}$$

Answer

The component of the electric field normal to the flat surface is constant over the surface

Answer

The circumference of the flat surface is an equipotential

Electrostatics & Capacitance

A point charge $+Q is placed just outside an imaginary hemispherical surface of radius R$ as shown in the figure. Which of the following statements is/are correct?

Choose the correct answer: