Electric Potential MCQ Quiz in मल्याळम - Objective Question with Answer for Electric Potential - സൗജന്യ PDF ഡൗൺലോഡ് ചെയ്യുക

CONCEPT: 

• The amount of work done in moving a unit positive charge in an electric field from infinity to that point without accelerating the charge against the direction of the electric field is electrostatic potential.
• The potential due to a point charge is given by

\(\Rightarrow V = \frac{1}{4\pi\epsilon_{0}} \frac{q}{r}\)

Where q = Charge, r = Distance, V = Electrostatic potential 

EXPLANATION : 

Let, V = The electrostatic potential when the first charge is brought from infinity to r

U = The electrostatic potential when the second charge is brought from infinity to \(\frac{r}{2}\)

• The value of electrostatic potential V is given by

\(\Rightarrow V = \frac{1}{4\pi \epsilon_{0}} \frac{Q}{r}\)    ------(1)

• The value of electrostatic potential U is given by

\(\Rightarrow U = \frac{1}{4\pi \epsilon_{0}} \frac{Q}{\frac{r}{2}}\)

\(\Rightarrow U = 2\times \frac{1}{4\pi \epsilon_{0}} \frac{Q}{r}\)          \([\because V = \frac{1}{4\pi \epsilon_{0}} \frac{Q}{r}]\)

\(\Rightarrow U = 2\times V\)       --------(2)

The difference in equation 2 and 1 will give you the potential difference and is 

\(\Rightarrow \Delta V = U - V = 2 V - V = V\)

• Hence option 2 is the answer

Gauss's Law of electrostatics: 

According to gauss’s law, the total electric flux through a closed surface enclosing a charge is 1/ϵ0 times the magnitude of the charge enclosed.

\({ϕ _{electric}} = \frac{Q}{{{ϵ_0}}}\)

\(\oint \vec{E}\cdot d\vec{S}=\frac{{{Q}_{in}}}{{{ϵ }_{0}}}\)

Where:

Φ = electric flux

Qin = charge enclosed by gaussian surface

ϵ0 = permittivity of space (8.85 × 10-12 C2/Nm2)

dS = surface area

Gaussian Surface:

 A Gaussian surface is a closed surface in 3D space through which the flux of a vector field is calculated.

1) It is a closed surface used with Gauss's law for the corresponding field by performing a surface integral, in order to calculate the total amount of the source quantity enclosed.

2) It is a surface vector used to calculate Gauss's flux.

3) It can be of any shape and size.

Electric potential:

Electric potential is equal to the amount of work done per unit charge by an external force to move the charge q from infinity to a specific point in an electric field.

⇒ \(V=\frac{W}{q}\)

Potential due to a single charged particle Q at a distance r from it is given by:

⇒ \( V=\frac{Q}{4\piϵ_{0}r}=E× r\)

Where, 

ϵ0 is the permittivity of free space and has a value of 8.85 × 10-12 F/m in SI units

Application:

By gauss law,

\(\oint \vec{E}\cdot d\vec{S}=\frac{{{Q}_{in}}}{{{ϵ }_{0}}}\)

E .A = q/ϵ

Here, A = Area of gaussian surface

But as we have to find the electric field in between r₁ and r₂ at any point, So we take the minimum radius i.e r₁ and A becomes 4πr₁²

Hence,

E × 4πr12  = q/ϵ          

\(E = \frac{q}{4\pi \epsilon r_1^2}\)

Therefore,

Electric potential is given by V = E× r₁

\(V = \dfrac{q\times r_1}{4\pi \epsilon r_1^2}\)

\(V = \dfrac{q}{4\pi \epsilon r_1}\)

CONCEPT: 

Electrostatic potential energy

• The potential energy of charge q at a point is the work done by the external force in bringing the charge q from infinity to that point.
• The electrostatic potential is given by

\(⇒ \Delta W = \int_{P}^{R} F.dr\)

\(⇒ \Delta U = W\)

Where F = Force W = Work, U = Electrostatic potential energy 

EXPLANATION: 

• In bringing a charge from infinity to point an external force that is just equal to electrostatic repulsive force must be applied. 

⇒ F_ext = F_{Electrostatic} 

• Hence option 1 is the answer 
• Nuclear, gravitational, wander walls forces are not related to electrostatics. Hence options 2, 3, and 4 are incorrect.

Concept:

Electric Potential: It is the amount of work needed to move a unit charge from a reference point to a specific point against the electric field.

If the two points are specified then the total work done in moving the charge from one point to another with respect to the reference point charge is given as,

\(W=-\frac{1}{4\pi {{\epsilon }_{0}}}{{q}_{1}}{{q}_{2}}\left( \frac{1}{{{r}_{1}}}-\frac{1}{{{r}_{2}}} \right)=k{{q}_{1}}{{q}_{2}}\left( \frac{1}{{{r}_{2}}}-\frac{1}{{{r}_{1}}} \right)\)

Where, k = 9 × 109 N-m2 C-2

q1 and q2 = charges

r1 and r2 are distances

Electric Potential due to point charge: Consider a point charge is placed at the distance of ‘r’ from the charge ‘Q’ then the potential due to charge is,

\({{V}_{p}}=\frac{kQ}{r}\)

Calculation:

Given, q₁ = 1 μC, q₂ = 10 μC

Let r₁ be the distance the two-point charges one placed placed at (0, 0, 0) and other at (0, 10 , 0)

r₁ = 10 m

Let r₂ be the final distance between the two-point charges one placed at origin (0, 0, 0) and other at (5, 0, 0)

r₂ = 5 m

Total work done in moving the pont charge from (0, 0, 0) to (5, 0, 0) is

\(W=k{{q}_{1}}{{q}_{2}}\left( \frac{1}{{{r}_{2}}}-\frac{1}{{{r}_{1}}} \right)\)

\(W=9\times 10^9\times 10^{-11}\left( \frac{1}{5}-\frac{1}{10} \right)\)

W = 9 mJ

Concept:

P.E of the field = 0 for equilibrium

Calculation:

\({V_1} = \frac{{2Q}}{{4\pi \epsilon 50}} + \frac{{{Q_3}}}{{4\pi \epsilon 100}}\) 

\({V_2} = \frac{Q}{{4\pi\epsilon \;50}} + \frac{Q}{{4\pi \epsilon\;100}}\) 

PE of the field:

Q₁V₁ + Q₂V₂ + Q₃V₃ = 0

\(\Rightarrow Q\left[ {\frac{{2Q}}{{4\pi \epsilon\;50}} + \frac{{{Q_3}}}{{4\pi\epsilon \;100}}} \right]\; + 2Q\) 

\(\left[ {\frac{Q}{{4\pi \epsilon\;50}} + \frac{{{Q_3}}}{{4\pi\epsilon \;50}}} \right] + {Q_3}\left[ {\frac{{2Q}}{{4\pi \epsilon\;50}} + \frac{Q}{{4\pi\epsilon \;100}}} \right] = 0\) 

\(\frac{{2Q}}{{4\pi \epsilon\;50}} + \frac{{{Q_3}}}{{4\pi\epsilon \;100}} + \frac{{2Q}}{{4\pi \epsilon\;50}}\) 

\( + \frac{{2{Q_3}}}{{4\pi \epsilon\;50}} + \frac{{2{Q_3}}}{{4\pi\epsilon \;50}} + \frac{{{Q_3}}}{{4\pi \epsilon\;100}} = 0\) 

\(\Rightarrow \frac{{4Q}}{{4\pi \epsilon\;50}} + \frac{{5{Q_3}}}{{4\pi \epsilon\;50}} = 0\) 

Q₃ = -0.8 Q

Concept:

The work done in moving a charge q between two points in an electric field can be calculated using the formula:

W = q * (V_M - VN)

Where W is the work done, q is the charge, and V_M and V_N are the electric potentials at points M and N, respectively.

Since the points M and N are at the same distance from the charge Q, the potential at both points is the same:

V_M = V_N = 1 / (4πε₀) * Q / r

Thus, the difference in potential between the two points is:

V_M - V_N = 0

The work done in moving the charge between points M and N is given by:

W = (V_M - V_N) * q = 0

∴ The work done in moving the charge is 0 J.

Concept:

According to Gauss’s law:

\(\oint \vec D \cdot \overrightarrow {ds} \) = charge enclosed

In point or Differential form:

\(\nabla \cdot \vec D = t\) 

And, \(\vec E = - \nabla \phi \) 

Calculation:

ϕ = 4x² + 3y² + Cz²

∇ϕ = 8x + 6y + 2Cz

\(\vec E = - 8x - 6y - 2cz\) 

\(\nabla \cdot \vec E = - 8 - 6 - 2C\) 

For a source-free region:

t = 0

∴ \(\nabla \cdot \vec D = 0\) 

\(\nabla \cdot \vec E = 0\) 

-8 – 6 – 2C = 0

2C = -14

C = -7

The potential difference gives the amount of work done in bringing a unit charge from one place to another.

Given, E = (xu_x + yu_y + zu_y) Volt/m

Let the potential difference between x and y be V_xy

The amount of work done in bringing a unit positive charge from Y(1, 2, 3) to X(2, 0, 0) will be:

\(= -\mathop \smallint \limits_{\left( {1,2,3} \right)}^{\left( {2,0,0} \right)} \bar E.\overline {dl} \)

\(= -\mathop \smallint \limits_{\left( {1,2,3} \right)}^{\left( {2,0,0} \right)} \left( {xdx + ydy + zdy} \right)\)

\(= -\frac{1}{2}{\rm{[}}{x^2}{\rm{|}}_1^2 + \left. {{y^2}} \right|_2^0 + \left. {{z^2}} \right|_3^0]\;\)

\(= -\frac{1}{2}\left[ {3 - 4 - 9} \right]\)

= 5 V

CONCEPT: 

• The amount of work done in moving a unit positive charge in an electric field from infinity to that point without accelerating the charge against the direction of the electric field is electrostatic potential.
• The potential due to a point charge is given by

\(\Rightarrow V = \frac{1}{4\pi\epsilon_{0}} \frac{q}{r}\)

Where q = Charge, r = Distance, and V = Electrostatic potential 

• The potential at the common center of two concentric spheres of radius r and R with Q amount of charges distributed on it is given by  [ Assume both the sphere have the same surface charge density ]

\(\Rightarrow V = \frac{KQ(r+R)}{(R^{2}+r^{2})}\)

Where Q = Charge, r = radius of the smaller sphere R = Radius of the larger sphere, \(K =\frac{1}{4\pi \epsilon_{0}} = 9\times 10^{9}\)

CALCULATION : 

Given - r = 4 cm R = 8 cm , Q = 12 μC

• The potential at the center of the two-sphere is 

\(\Rightarrow V = \frac{kQ(r+R)}{(R^{2}+r^{2})}\)

Substituting the given values in the above equation

\(\Rightarrow V = \frac{9\times 10^{9}\times 12\times 10^{-6}\times (4\times 10^{-2}+8\times 10^{-2})}{((8\times 10^{-2})^{2}+(4\times 10^{-2})^{2})}\)

\(\Rightarrow V = \frac{9\times 12\times12 \times 10^{3}}{8\times 10^{-3}} = 162\times 10^{4} V\)

• Hence option 4 is the answer

Concept:

The potential point charge at a distance r is defined as the amount of work done in moving that charge from infinite to the point from where it is calculated.

\(V = \frac{W}{Q}\)

W = work done

Q = Charge

And W = F.d

F = force

d = distance between point and charge = r

\(F = \frac{1}{{4\pi {\epsilon_0}}}\frac{{{Q_1}{Q_2}}}{{{r^2}}}\)

\( \Rightarrow W = \frac{1}{{4\pi {\epsilon_0}}}\frac{{{Q_1}{Q_2}}}{r}\)

\( \Rightarrow V = \frac{1}{{4\pi {\epsilon_0}}}\frac{Q}{r}\)

Calculation:

    \(V = {9 * 10^9 *5*10^{-6} \over 2}\)

        =22.5 KV