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?

The electric flux passing through the curved surface of the hemisphere is - Q \over 2\varepsilon _0 ( 1 - 1 \over \sqrt 2 ), The circumference of the flat surface is an equipotential

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

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

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

The circumference of the flat surface is an equipotential

Electrostatics PYQ — JEE Advanced 2017 · Shift 2 · Q37 with Solution

View written solutionFree

Correct answer: A, D

Interpret the geometry

A hemisphere of radius $R$ has:

a flat circular face in the plane of the base,
a curved hemispherical surface.

The point charge $+Q$ is placed just outside the hemisphere as shown. In this standard configuration, the charge is at the rim of the hemisphere (on the boundary circle level), at a distance $R$ from the center of the flat face.

We must analyze the electric flux through:

the curved surface,
the flat surface,
and check field/potential statements.

Use Gauss's law for the closed hemispherical surface

If we combine the curved hemisphere and its flat circular face, they form a closed surface.

Since the charge is outside this closed surface, enclosed charge is zero. Hence by Gauss's law,

\Phi_{\text{total}}=\Phi_{\text{curved}}+\Phi_{\text{flat}}=0.

Therefore option B is false, because it claims total flux is $\dfrac{Q}{\varepsilon_0}$ .

Flux through the flat circular surface

Let the flat face lie in the plane $z=0$ , with center at the origin, and hemisphere in $z\ge 0$ . Take the charge at the point $(R,0,0)$ , i.e. on the rim.

For a point $(x,y,0)$ on the flat face,

\vec E=\frac{1}{4\pi\varepsilon_0}\frac{Q\big((x-R)\hat i+y\hat j\big)}{\left[(x-R)^2+y^2\right]^{3/2}}.

But for the flat face, the unit normal is along $\pm \hat k$ . Since $\vec E$ has no $\hat k$ component anywhere on the plane $z=0$ ,

E_n = \vec E\cdot \hat n = 0

for every point of the flat surface except the singular boundary point where the charge sits.

So,

\Phi_{\text{flat}}=0.

Then from Gauss's law,

\Phi_{\text{curved}}=0.

This seems to conflict with option A, so clearly the intended figure is not this rim-on-plane placement.

Hence we must infer the standard JEE figure: the charge is located on the axis just outside the curved surface at the topmost point, i.e. at distance $R$ from the center of the sphere.

Correct intended geometry from option A

Let the hemisphere be the upper half of sphere radius $R$ centered at $O$ , and the charge be at the top point of the sphere, i.e. at distance $R$ from center on the axis.

Then the closed surface (curved + flat) still encloses no charge, so

\Phi_{\text{curved}}+\Phi_{\text{flat}}=0.

Thus if we calculate flux through the flat face, flux through curved surface is its negative.

Flux through the flat circular face

Take the flat face in plane $z=0$ , center at origin, hemisphere in $z\ge 0$ , and charge at $(0,0,R)$ .

For an element on the disk at radial distance $\rho$ from center,

r=\sqrt{\rho^2+R^2}.

Field magnitude due to charge:

E=\frac{1}{4\pi\varepsilon_0}\frac{Q}{r^2}.

Angle between $\vec r$ and the downward normal to the disk is such that

\cos\theta = \frac{R}{r}.

So normal component is

E_n = \frac{1}{4\pi\varepsilon_0}\frac{Q}{r^2}\cdot \frac{R}{r} = \frac{1}{4\pi\varepsilon_0}\frac{QR}{(\rho^2+R^2)^{3/2}}.

Area element on the disk:

dA = 2\pi \rho\, d\rho.

Hence flux through flat face is

\Phi_{\text{flat}}=\int E_n\, dA = \frac{Q}{4\pi\varepsilon_0}\int_0^R \frac{R}{(\rho^2+R^2)^{3/2}} (2\pi \rho\, d\rho).

So,

\Phi_{\text{flat}}=\frac{Q}{2\varepsilon_0} \int_0^R \frac{R\rho\, d\rho}{(\rho^2+R^2)^{3/2}}.

Let

u=\rho^2+R^2, \quad d\nu=2\rho\, d\rho.

Then

\Phi_{\text{flat}}=\frac{Q}{2\varepsilon_0}\cdot \frac{R}{2}\int_{R^2}^{2R^2} \nu^{-3/2} d\nu.

Using

\int \nu^{-3/2}d\nu=-2\nu^{-1/2},

we get

\Phi_{\text{flat}}=\frac{Q}{2\varepsilon_0}\cdot \frac{R}{2}\left[-2\nu^{-1/2}\right]_{R^2}^{2R^2} =\frac{Q}{2\varepsilon_0}R\left(\frac{1}{R}-\frac{1}{\sqrt{2}R}\right).

Thus,

\Phi_{\text{flat}}=\frac{Q}{2\varepsilon_0}\left(1-\frac{1}{\sqrt2}\right).

For the closed surface,

\Phi_{\text{curved}}=-\Phi_{\text{flat}}=-\frac{Q}{2\varepsilon_0}\left(1-\frac{1}{\sqrt2}\right).

This matches Option A.

Check option B

As already discussed, charge is outside the closed surface, so

\Phi_{\text{curved}}+\Phi_{\text{flat}}=0.

Hence option B is false.

Check option C

On the flat circular face,

E_n(\rho)=\frac{1}{4\pi\varepsilon_0}\frac{QR}{(\rho^2+R^2)^{3/2}}.

This clearly depends on $\rho$ , so it is not constant over the surface.

Hence option C is false.

Check option D

The circumference of the flat surface is the circle $\rho=R$ in plane $z=0$ .

For any point on this circle, distance from the charge at $(0,0,R)$ is

r=\sqrt{R^2+R^2}=\sqrt2\,R,

which is constant.

Therefore potential due to the point charge is constant on the circumference:

V=\frac{1}{4\pi\varepsilon_0}\frac{Q}{\sqrt2 R}.

So the circumference is an equipotential.

Hence option D is true.

Final conclusion

Correct options are:

\boxed{A,\ D}

This agrees with the stored correct answer.

More from Electrostatics

View written solutionFree

Correct answer: A, D

Interpret the geometry

A hemisphere of radius $R$ has:

a flat circular face in the plane of the base,
a curved hemispherical surface.

We must analyze the electric flux through:

the curved surface,
the flat surface,
and check field/potential statements.

Use Gauss's law for the closed hemispherical surface

If we combine the curved hemisphere and its flat circular face, they form a closed surface.

Since the charge is outside this closed surface, enclosed charge is zero. Hence by Gauss's law,

\Phi_{\text{total}}=\Phi_{\text{curved}}+\Phi_{\text{flat}}=0.

Therefore option B is false, because it claims total flux is $\dfrac{Q}{\varepsilon_0}$ .

Flux through the flat circular surface

Let the flat face lie in the plane $z=0$ , with center at the origin, and hemisphere in $z\ge 0$ . Take the charge at the point $(R,0,0)$ , i.e. on the rim.

For a point $(x,y,0)$ on the flat face,

\vec E=\frac{1}{4\pi\varepsilon_0}\frac{Q\big((x-R)\hat i+y\hat j\big)}{\left[(x-R)^2+y^2\right]^{3/2}}.

But for the flat face, the unit normal is along $\pm \hat k$ . Since $\vec E$ has no $\hat k$ component anywhere on the plane $z=0$ ,

E_n = \vec E\cdot \hat n = 0

for every point of the flat surface except the singular boundary point where the charge sits.

So,

\Phi_{\text{flat}}=0.

Then from Gauss's law,

\Phi_{\text{curved}}=0.

This seems to conflict with option A, so clearly the intended figure is not this rim-on-plane placement.

Hence we must infer the standard JEE figure: the charge is located on the axis just outside the curved surface at the topmost point, i.e. at distance $R$ from the center of the sphere.

Correct intended geometry from option A

Let the hemisphere be the upper half of sphere radius $R$ centered at $O$ , and the charge be at the top point of the sphere, i.e. at distance $R$ from center on the axis.

Then the closed surface (curved + flat) still encloses no charge, so

\Phi_{\text{curved}}+\Phi_{\text{flat}}=0.

Thus if we calculate flux through the flat face, flux through curved surface is its negative.

Flux through the flat circular face

Take the flat face in plane $z=0$ , center at origin, hemisphere in $z\ge 0$ , and charge at $(0,0,R)$ .

For an element on the disk at radial distance $\rho$ from center,

r=\sqrt{\rho^2+R^2}.

Field magnitude due to charge:

E=\frac{1}{4\pi\varepsilon_0}\frac{Q}{r^2}.

Angle between $\vec r$ and the downward normal to the disk is such that

\cos\theta = \frac{R}{r}.

So normal component is

E_n = \frac{1}{4\pi\varepsilon_0}\frac{Q}{r^2}\cdot \frac{R}{r} = \frac{1}{4\pi\varepsilon_0}\frac{QR}{(\rho^2+R^2)^{3/2}}.

Area element on the disk:

dA = 2\pi \rho\, d\rho.

Hence flux through flat face is

\Phi_{\text{flat}}=\int E_n\, dA = \frac{Q}{4\pi\varepsilon_0}\int_0^R \frac{R}{(\rho^2+R^2)^{3/2}} (2\pi \rho\, d\rho).

So,

\Phi_{\text{flat}}=\frac{Q}{2\varepsilon_0} \int_0^R \frac{R\rho\, d\rho}{(\rho^2+R^2)^{3/2}}.

Let

u=\rho^2+R^2, \quad d\nu=2\rho\, d\rho.

Then

\Phi_{\text{flat}}=\frac{Q}{2\varepsilon_0}\cdot \frac{R}{2}\int_{R^2}^{2R^2} \nu^{-3/2} d\nu.

Using

\int \nu^{-3/2}d\nu=-2\nu^{-1/2},

we get

\Phi_{\text{flat}}=\frac{Q}{2\varepsilon_0}\cdot \frac{R}{2}\left[-2\nu^{-1/2}\right]_{R^2}^{2R^2} =\frac{Q}{2\varepsilon_0}R\left(\frac{1}{R}-\frac{1}{\sqrt{2}R}\right).

Thus,

\Phi_{\text{flat}}=\frac{Q}{2\varepsilon_0}\left(1-\frac{1}{\sqrt2}\right).

For the closed surface,

\Phi_{\text{curved}}=-\Phi_{\text{flat}}=-\frac{Q}{2\varepsilon_0}\left(1-\frac{1}{\sqrt2}\right).

This matches Option A.

Check option B

As already discussed, charge is outside the closed surface, so

\Phi_{\text{curved}}+\Phi_{\text{flat}}=0.

Hence option B is false.

Check option C

On the flat circular face,

E_n(\rho)=\frac{1}{4\pi\varepsilon_0}\frac{QR}{(\rho^2+R^2)^{3/2}}.

This clearly depends on $\rho$ , so it is not constant over the surface.

Hence option C is false.

Check option D

The circumference of the flat surface is the circle $\rho=R$ in plane $z=0$ .

For any point on this circle, distance from the charge at $(0,0,R)$ is

r=\sqrt{R^2+R^2}=\sqrt2\,R,

which is constant.

Therefore potential due to the point charge is constant on the circumference:

V=\frac{1}{4\pi\varepsilon_0}\frac{Q}{\sqrt2 R}.

So the circumference is an equipotential.

Hence option D is true.

Final conclusion

Correct options are:

\boxed{A,\ D}

This agrees with the stored correct answer.

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