Physics M.C.Q [1M]

3. W_A = W_B = W_C

Explanation:

Electric Potential at A 'due to q' þ $\text{V}_\text{A}=\frac{\text{Kq}}{\text{r}}$

Electric Potetial at B 'due to q' þ $\text{V}_\text{B}=\frac{\text{Kq}}{\text{r}}$

& Electric potential at c 'due to q' þ $\text{V}_\text{C}=\frac{\text{Kq}}{\text{r}}$

Work done $=-\text{D}_\text{u}=-\text{q}\text{DV}$ {Let at 'P', V_p = 0}

Here $\text{V}_\text{A}=\text{V}_\text{B}=\text{V}_\text{C}$

The work done is taking a point charge from P to A, B & C is same.

So, $\text{W}_\text{A}=\text{W}_\text{B}=\text{W}_\text{C}$

2. May be equal to 20Vcm^-1
3. May be greater than 20Vcm^-1

Explanation:

$\triangle\text{V}=-\text{E}.\text{dr}$

$(\text{V}_\text{f}-\text{V}_\text{i})=-\text{E}.\text{dr}=\text{E}_\text{x}(\text{B}-\text{A})$

$(80-120)=-\text{E}_\text{x}.(2)$

$\text{E}_\text{x}=\frac{40}{2}=20\text{v/m}$

If electric field lines lies in 'x ' direction than it may be equal to 20v/cm.

If Electric field lines liesin 'x - y' direction than it may be greater than 20v/cm.

2. If V = 0, E must be zero.

Explantion:

Electric field is negative gradient of electric potential.

$\text{E}=-\text{grad}(\text{V})$

$\text{E}=-\frac{\text{dV}}{\text{dr}}$

If $\text{E}=0$

$-\frac{\text{dV}}{\text{dr}}=0$

This implies

V = constant.

A constant can be zero or a quantifiable number.

4. May increase or decrease.

Explanation:

When the separation between two charges is increased, the electric potential Energy of charge may incease or decrease.

If Both charge are like charge then electric potential energy of charge decreases.

$\text{U}=\frac{\text{k}\text{q}_1\text{q}_2}{\text{r}}$

If Both charge are unlike charge then electric potential energy of charge increases.

$\text{U}=\frac{-\text{kq}_1\text{q}_2}{\text{r}}$

3. Potential energy of a two-charge system.
4. Change in potential energy of a two-charge system.

Explanation:

$\text{V}_\text{p}=\frac{\text{KQ}}{\text{r}}-\frac{\text{KQ}}{\text{r}}=0$

1. Continuously increases.

Explanation:

$\text{V}=\frac{\text{KQ}}{\text{r}}$

V → Electric Potential

$\text{V}_\text{p}=\frac{\text{KQ}}{\text{x}}+\frac{\text{KQ}}{\text{r}-\text{x}}=\frac{\text{KQ}\text{r}}{\text{x}(\text{r}-\text{x})}$

$\frac{\text{dvp}}{\text{dx}}=\frac{-\text{kQr}(\text{r}-2\text{x})}{\big(\text{r}(\text{r}-\text{x})\big)^2}=0$

$\text{r}=2\text{x},\ \text{x}=\frac{\text{r}}{2}$

$\text{V}_\text{Pmin}=\frac{\text{KQ}}{\frac{\text{r}}{2}}+\frac{\text{KQ}}{\frac{\text{r}}{2}}=\frac{4\text{KQ}}{\text{r}}$ at $\Big(\text{x}=\frac{\text{r}}{2}\Big)$

1. Zero.

Explanation:

In a circle Electric field due tot 'Q' is always perpendicular to the displacement oifthe charge 'q'.

Workdone $=\vec{\text{F}}.\vec{\text{d}}$

$=\big(\text{q}\vec{\text{E}}\big).\vec{\text{d}}=\text{q}\big(\vec{\text{E}}.\vec{\text{d}}\big)=0$

$\therefore\vec{\text{E}}\ \bot\ \vec{\text{d}}$

Workdoen by the Electric field on the rotating charge in one complete revolution is zero.

3. It is proportional to r².

Explanation:

$\text{DV}=-\text{E}.\text{dr}$

Given $(\text{E}\propto\text{r})$

$(\text{V}-0)=-\text{E}.\text{dr}$ 

$þ\text{E}\propto\text{r}^2$

1. Increases.

Explanation:

Electric Potential Energy $=\text{q}\text{Dv}$

$=\text{q}(\text{v}_\text{f}-\text{v}_\text{i})$

If positive charge is shifted from a Low potential region to a High-Potential region, then electric Potential Energy increases.

1. A

Explanation:

$\text{E}=\text{E}_0\hat{\text{i}}$ (Given)

(Potential at A):-

$\text{V}_\text{A}=\big(\text{E}_0\hat{\text{i}}\times\text{a}\hat{\text{i}}\big)=\text{E}_09$

(Potential atB):-

$\text{V}_\text{B}=\big(\text{E}_0\hat{\text{i}}\times\text{a}\hat{\text{i}}\big)=0$

(Potential at C):-

$\text{V}_\text{C}=\big(\text{E}_0\hat{\text{i}}\times(-\text{a})\text{i}\big)=\text{E}_0\text{a}$

(Potential atD):-

$\text{V}_\text{D}=(\text{E}_0\hat{\text{i}}\times(-\text{a})\hat{\text{j}})=0$

4. The torque on the dipole due to the field may be zero.

Explanation:

In the uniform Electric field the net electric force on the dipole is alwas zero.

In uniform Electric field the torque on the dipole due to field may be zero.

T = 0

Here t Þ Torque

$\text{T}=2\text{qEl}\sin\text{q }\otimes\neq0$

2. The magnitudes of the forces will be equal.

Explanation:

Proton contain positive charge = 1.6 × 10^-19 = e

Electron contain negative charge = -1.6 × 10^-19 = -e

Force inoniforma electric field is = qE

$\therefore\ $The Magniludes of Electric force is equal But direction of electric force is opposite to each other.

mass of proton = 1.67 × 10^-27Kg.

mass of electron = 9.1 × 10^-31Kg.

So that Magnitudes of their acceleration will be unequal.