Tag: applications of gauss's law

Questions Related to applications of gauss's law

The magnitude of the electric field on the surface of a sphere of radius $r$ having a uniform surface charge density $\sigma$ is

applications of gauss's law coulomb's law physics

  1. $\sigma / \epsilon _{0}$

  2. $\sigma / 2\epsilon _{0}$

  3. $\sigma / \epsilon _{0}r$

  4. $\sigma / 2\epsilon _{0}r$

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Correct Option: A
Explanation:
The magnitude of the electric field on the surface of radius $=r$
Charge density $=6$
Then, $E=\dfrac { 6 }{ { \epsilon  } _{ 0 } } $
The electric field is independent of the surface radius.

Consider a thin spherical shell of radius $R$ consisting of uniform surface charge density $\sigma$. The electric field at a point of distance $x$ from its centre and outside the shell is

applications of gauss's law coulomb's law physics

  1. inversely proportional to $\sigma$

  2. directly proportional to ${x}^{2}$

  3. directly proportional to $R$

  4. inversely proportional to ${x}^{2}$

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Correct Option: D
Explanation:
For a thin uniformly charged spherical shell, the field points outside the shell at a distance $x$ from the centre is
$E=\cfrac { 1 }{ 4\pi { \varepsilon  } _{ 0 } } \cfrac { Q }{ { x }^{ 2 } } $
If the radius of the sphere is $R,Q=\sigma 4\pi { R }^{ 2 }$
$\therefore E=\cfrac { 1 }{ 4\pi { \varepsilon  } _{ 0 } } \cfrac { \sigma 4\pi { R }^{ 2 } }{ { x }^{ 2 } } =\cfrac { \sigma { R }^{ 2 } }{ { { \varepsilon  } _{ 0 }x }^{ 2 } } $
This is inversely proportional to square of the distance from the centre. It is as if the whole charge is concentrated at the centre

Two charged spheres having radii a and b are joined with a wire then the ratio of electric field $\dfrac{E _a}{E _b}$ on their surface is?

applications of gauss's law coulomb's law physics

  1. a/b

  2. b/a

  3. ba

  4. None of these

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Correct Option: B
Explanation:

When the two spheres are connected by a wire, then both of them acquire the same potential say $V$.


We also know that the electric field on the surface of a sphere $E=\dfrac{Q}{4\pi\epsilon _o r^2}$
and potential on the surface is given by $V=\dfrac{Q}{4\pi\epsilon _or}$

$\implies E=\dfrac{V}{r}$

Here, V is constant , hence  $E\propto \dfrac{1}{r}$

$\implies \dfrac{E _a}{E _b}=\dfrac{b}{a}$

Charges $Q _1$ and $Q _2$ are placed inside and outside respectively of an uncharged conducting shell. Their seperation is r.

applications of gauss's law coulomb's law physics

  1. The force on $Q _1$ is zero.

  2. The force on $Q _1$ is $\displaystyle k \frac{Q _1 Q _2}{r^2}$

  3. The force on $Q _2$ is $\displaystyle k \frac{Q _1 Q _2}{r^2}$

  4. The force on $Q _2$ is zero.

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Correct Option: A,C
Explanation:

As the electric field inside the conducting shell is zero , so the force on the inner charge, $Q _1$ will be zero.
The electric field at outside charge $Q _2$ due to $Q _1$ is $E=k\frac{Q _1}{r^2}$
Force on $Q _2$ is $F=Q _2E=k\frac{Q _1Q _2}{r^2}$