Physics MCQs for NEET — Practice Questions with Answers

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The potential energy of a single charge 'q' at a point 'r' in an external electric field is given by:

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Explanation

The NCERT states, 'Potential energy of q at r in an external field = qV(r) where V(r) is the external potential at the point r.' (Eq. 2.27 and accompanying text).

Which of the following statements is true regarding the influence of a charge 'q' on the external sources producing the electric field?

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Explanation

The provided text mentions, 'We assume that the charge q does not significantly affect the sources producing the external field. This is true if q is very small, or the external sources are held fixed by other unspecified forces.' (Section 2.8).

If an electron with charge $q = e = 1.6 \times 10^{-19} C$ is accelerated by a potential difference of 1 Volt, what is the energy gained in Joules?

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Explanation

The NCERT states, 'if an electron with charge $q = e = 1.6 \times 10^{-19} C$ is accelerated by a potential difference of $\Delta V = 1$ volt, it would gain energy of $q\Delta V = 1.6 \times 10^{-19} J$.' (Section 2.8.1).

What is the equivalent energy value of 1 MeV in Joules?

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Explanation

The NCERT specifies energy units: '1 MeV = $10^6 eV = 1.6 \times 10^{-13} J$.' (Section 2.8.1).

For a system of two charges $q_1$ and $q_2$ located at $r_1$ and $r_2$ respectively, in an external field $V(r)$, the total potential energy of the system is given by:

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Explanation

The total potential energy of the system is the sum of the work done in bringing each charge into the external field and the work done against the field of the other charge. The NCERT states, 'Potential energy of the system = the total work done in assembling the configuration $= q_1V(r_1) + q_2V(r_2) + \frac{1}{4\pi\epsilon_0} \frac{q_1q_2}{r_{12}}$' (Eq. 2.29).

When bringing a charge $q_2$ from infinity to a point $r_2$ in an external field and in the presence of another charge $q_1$ (already at $r_1$), the work done involves contributions from:

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Explanation

The NCERT text explains: 'Next, we consider the work done in bringing $q_2$ to $r_2$. In this step, work is done not only against the external field E but also against the field due to $q_1$.' (Section 2.8.2).

A dipole with charges $+q$ and $-q$ is placed in a uniform electric field $E$. If the dipole moment $\vec{p}$ is perpendicular to $\vec{E}$ (i.e., $\theta = \pi/2$), and the potential energy is chosen to be zero at this angle, what is the potential energy when the dipole makes an angle $\theta$ with the field?

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Explanation

For a dipole in an external field, the potential energy is given by $U(\theta) = -pE \cos\theta$. This formula is derived by integrating the work done from an initial $\theta_0 = \pi/2$ (where $U=0$) to a final $\theta$. The NCERT states, 'We can then write, $U(\theta) = -pE \cos\theta$' (Eq. 2.32).

In a uniform electric field, an electric dipole experiences:

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Explanation

The NCERT states, 'As seen in the last chapter, in a uniform electric field, the dipole experiences no net force; but experiences a torque $\vec{\tau} = \vec{p} \times \vec{E}$.' (Section 2.8.3).

Consider a system of two charges $+7 \mu C$ and $-2 \mu C$ with no external field, placed at $(-9 cm, 0, 0)$ and $(9 cm, 0, 0)$ respectively. What is their electrostatic potential energy?

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Explanation

Using the formula for potential energy of two charges $U = \frac{1}{4\pi\epsilon_0} \frac{q_1q_2}{r_{12}}$. Here $q_1 = 7 \times 10^{-6} C$, $q_2 = -2 \times 10^{-6} C$, and $r_{12} = 9 cm - (-9 cm) = 18 cm = 0.18 m$. $U = (9 \times 10^9) \frac{(7 \times 10^{-6})(-2 \times 10^{-6})}{0.18} = 9 \times 10^9 \times \frac{-14 \times 10^{-12}}{0.18} = \frac{-126 \times 10^{-3}}{0.18} = -0.7 J$. This matches the Example 2.5(a) in the NCERT.

If the system of charges from the previous question ( $+7 \mu C$ and $-2 \mu C$ ) is now placed in an external electric field $E = A(1/r^2)$ where $A = 9 \times 10^5 NC^{-1} m^2$, what additional energy contribution needs to be considered for the total electrostatic energy compared to the case with no external field?

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Explanation

The NCERT states for a system of charges in an external field, the total potential energy includes the mutual interaction energy of the charges plus the energy of interaction of each charge with the external electric field. For two charges, this additional part is $q_1V(r_1) + q_2V(r_2)$. (Example 2.5(c) and equation 2.29).

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