Tag: applications of gauss's law

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Multiple choice applications of gauss's law coulomb's law physics

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

  1. $\sigma / \epsilon _{0}$
  2. $\sigma / 2\epsilon _{0}$
  3. $\sigma / \epsilon _{0}r$
  4. $\sigma / 2\epsilon _{0}r$
Reveal answer Fill a bubble to check yourself
A Correct answer
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.
Multiple choice applications of gauss's law coulomb's law physics

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

  1. inversely proportional to $\sigma$
  2. directly proportional to ${x}^{2}$
  3. directly proportional to $R$
  4. inversely proportional to ${x}^{2}$
Reveal answer Fill a bubble to check yourself
D Correct answer
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
Multiple choice applications of gauss's law coulomb's law physics

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?

  1. a/b

  2. b/a

  3. ba

  4. None of these
Reveal answer Fill a bubble to check yourself
B Correct answer
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  $E\propto \dfrac{1}{r}$

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

Multiple choice applications of gauss's law coulomb's law physics

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

  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.
Reveal answer Fill a bubble to check yourself
A,C Correct answer
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 $F=Q _2E=k\frac{Q _1Q _2}{r^2}$