Tag: coulomb's law

Quizzes (4) Questions (102)

Questions Related to coulomb's law

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 $F=Q _2E=k\frac{Q _1Q _2}{r^2}$

Multiple choice physics coulomb's law field strength and potential gradient electric field as gradient of potential relation between electric field and electric potential

If we move in a direction opposite to the electric lines of force:

  1. electrical potential decreases

  2. electrical potential increases

  3. electrical potential remains uncharged

  4. nothing can be said.
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Electric field lines point from higher to lower potential. Moving in the direction opposite to the lines of force means moving toward higher potential.

Multiple choice physics coulomb's law field strength and potential gradient electric field as gradient of potential relation between electric field and electric potential

A uniform wire $10 \,cm$ long is carrying a steady current. The potential drop across it is $10V$. The electric field inside it is _____

  1. zero
  2. $1Nm^{-1}$
  3. $10 \,Vm^{-1}$
  4. $100 \,Vm^{-1}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

$E = \dfrac{V}{r} = \dfrac{10V}{10 \times 10^{-2}m} = \dfrac{10 \times 100}{10} = 100 \dfrac{v}{m}$

Multiple choice physics coulomb's law field strength and potential gradient electric field as gradient of potential relation between electric field and electric potential

The electric potential while moving along the lines of force

  1. decreases

  2. increases

  3. remains constant

  4. becomes infinite
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Electric field lines are defined as the path along which a positive test charge would move. Since a positive charge moves from higher to lower potential, the potential decreases along the lines of force.

Multiple choice physics coulomb's law field strength and potential gradient electric field as gradient of potential relation between electric field and electric potential

$E=-\dfrac{dV}{dr}$, here negative sign signified that

  1. E is opposite to V

  2. E is negative

  3. E increases when V decreases

  4. E is directed in the direction of decreasing V
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The negative sign is just a convention and it signifies that the direction of E is opposite to the direction in which potential increases.

Multiple choice physics coulomb's law field strength and potential gradient electric field as gradient of potential relation between electric field and electric potential

The ratio of electric force $ ( F _e ) $ to gravitational force acting between two electrons will be:

  1. $ 1 \times 10^{36} $
  2. $ 2 \times 10^{39} $
  3. $ 2.5\times 10^{39} $
  4. $ 3 \times 10^{39} $
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The ratio of electrostatic force to gravitational force between two electrons is approximately 4.17 * 10^42. The provided options are all in the 10^39 range, which is a common textbook approximation for this ratio.

Multiple choice physics coulomb's law field strength and potential gradient electric field as gradient of potential relation between electric field and electric potential

Electric potential at ( x, y, z ) is given as $V$= $- x ^ { 2 } y \sqrt { z }$ Find the electrical field at (2 ,1, 1)

  1. $4 \hat { i } + 4 \hat { j } + 4 \hat { k }$
  2. $- 4 \hat { i } - 4 \hat { j } - 2 \hat { k }$
  3. $- 4 \hat { i } - 4 \hat { j } - 4 \hat { k }$
  4. $4 \hat { 1 } + 4 \hat { j } + 2 \hat { k }$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The electric field is the negative gradient of potential: E = -∇V. Computing partial derivatives: ∂V/∂x = -2xy√z, ∂V/∂y = -x²√z, ∂V/∂z = -x²y/(2√z). At point (2,1,1): Ex = -2(2)(1)(1) = -4, Ey = -(4)(1) = -4, Ez = -(4)(1)/(2×1) = -2. Therefore E = -(-4i - 4j - 2k) = 4i + 4j + 2k. The key is applying the gradient operator correctly and evaluating at the given point. Note that option D has a typo (should be î, not 1̂) but is clearly the intended answer.