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Physics MCQs for NEET — Practice Questions with Answers

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Ten electrons are equally spaced and fixed around a circle of radius R. Relative to V = 0 at infinity, the electrostatic potential V and the electric field E at the centre C are

$V \neq 0$ and $\vec{E} \neq 0$

$V \neq 0$ and $\vec{E} = 0$

V = 0 and $\vec{E} = 0$

V = 0 and $\vec{E} \neq 0$

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Explanation

At centre E = 0, V ≠ 0 because the electric field is a vector quantity but the electric potential is not a vector quantity.

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The displacement of a charge Q in the electric field $E = e_{1} \hat{i} + e_{2} \hat{j} + e_{3} \hat{k}$ is $\hat{r} = a \hat{i} + b \hat{j}$ . The work done is

$Q (a e_{1} + b e_{2})$

$Q \sqrt{{(a e_{1})}^{2} + {(b e_{2})}^{2}}$

$Q (e_{1} + e_{2}) \sqrt{a^{2} + b^{2}}$

$Q (\sqrt{e_{1}^{2} + e_{2}^{2})} (a + b)$

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Explanation

By using $W = Q (\vec{E} . Δ \vec{r})$

⇒ $W = Q [(e_{1} \hat{i} + e_{2} \hat{j} + e_{3} \hat{k}) . (a \hat{i} + b \hat{j})] = Q (e_{1} a + e_{2} b)$

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A cube of a metal is given a positive charge Q. For the above system, which of the following statements is true

Electric potential at the surface of the cube is zero

Electric potential within the cube is zero

Electric field is normal to the surface of the cube

Electric field varies within the cube

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Explanation

Electric lines of force are always normal to metallic body.

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Electric potential at any point is $V = - 5 x + 3 y + \sqrt{15} z$ , then the magnitude of the electric field is

$3 \sqrt{2}$

$4 \sqrt{2}$

$5 \sqrt{2}$

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Explanation

$E_{x} = - \frac{d V}{d x} = - (- 5) = 5;$ $E_{y} = - \frac{d V}{d y} = - 3$

and $E_{z} = - \frac{d V}{d z} = - \sqrt{15}$

$E_{n e t} = \sqrt{E_{x}^{2} + E_{y}^{2} + E_{z}^{2}} = \sqrt{{(5)}^{2} + {(- 3)}^{2} + {(- \sqrt{15})}^{2}} = 7$

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Kinetic energy of an electron accelerated in a potential difference of 100 V is

1.6 × 10^–17 J

1.6 × 10²¹ J

1.6 × 10^–29 J

1.6 × 10^–34 J

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Explanation

By using KE = QV ⇒ KE = 1.6 × 10^–19 × 100

= 1.6 × 10^–17 J

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If identical charges (–q) are placed at each corner of a cube of side b, then electric potential energy of charge (+q) which is placed at centre of the cube will be -

$\frac{8 \sqrt{2} q^{2}}{4 π ε_{0} b}$

$\frac{- 8 \sqrt{2} q^{2}}{π ε_{0} b}$

$\frac{- 4 \sqrt{2} q^{2}}{π ε_{0} b}$

$\frac{- 4 q^{2}}{\sqrt{3} π ε_{0} b}$

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Explanation

Length of the diagonal of a cube having each side b is $\sqrt{3} b .$ So distance of centre of cube from each vertex is $\frac{\sqrt{3} b}{2} .$

Hence potential energy of the given system of charge is

$U = 8 \times \{\frac{1}{4 π ε_{0}} . \frac{(- q) (q)}{\sqrt{3} b / 2}\} = \frac{- 4 q^{2}}{\sqrt{3} π ε_{0} b}$

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A proton is about 1840 times heavier than an electron. When it is accelerated by a potential difference of 1 kV, its kinetic energy will be -

1840 keV

1/1840 keV

1 keV

920 keV

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Explanation

KE = QV = e × 10³ V = 1KeV.

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A thin spherical conducting shell of radius R has a charge q. Another charge Q is placed at the centre of the shell. The electrostatic potential at a point p a distance $\frac{R}{2}$ from the centre of the shell is

$\frac{(q + Q)}{4 π ε_{0}} \frac{2}{R}$

$\frac{2 Q}{4 π ε_{0} R}$

$\frac{2 Q}{4 π ε_{0} R} - \frac{2 q}{4 π ε_{0} R}$

$\frac{2 Q}{4 π ε_{0} R} + \frac{q}{4 π ε_{0} R}$

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Explanation

For a conducting spherical shell, the electric field and potential inside the shell are due solely to the charge Q at the center. Outside the shell, the potential is the sum of the potentials due to Q and the charge q on the shell. At a distance R/2 from the center, the potential contributions from Q and q add up to give the given expression.

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A sphere of 4 cm radius is suspended within a hollow sphere of 6 cm radius. The inner sphere is charged to potential 3 e.s.u. and the outer sphere is earthed. The charge on the inner sphere is

54 e.s.u.

$\frac{1}{4}$ e.s.u.

30 e.s.u.

36 e.s.u.

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Four charges $+ Q, - Q, + Q, - Q$ are placed at the corners of a square taken in order. At the centre of the square

$E = 0, V = 0$

$E = 0, V \neq 0$

$E \neq 0, V = 0$

$E = 0, V \neq 0$

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