Practice free Electrostatics (Physics) NEET multiple-choice questions online with instant answers and detailed explanations. No login required.
For a point charge Q, if Q < 0, the work done by the external force in bringing a unit positive test charge from infinity to a point P is:
The NCERT text states: 'For Q < 0, V < 0, i.e., work done (by the external force) per unit positive test charge in bringing it from infinity to the point is negative.' (Page 49, Example 2.1 note).
The electric potential on the dipole axis for an electric dipole (where $\theta = 0$ or $\theta = \pi$) is given by:
The NCERT text states: 'From Eq. (2.15), potential on the dipole axis ($\theta = 0, \pi$) is given by $V = \pm \frac{1}{4\pi\epsilon_0} \frac{p}{r^2}$ (Eq. 2.16).' Here, for $\theta=0$, $\cos\theta=1$, so $V = \frac{1}{4\pi\epsilon_0} \frac{p}{r^2}$. For $\theta=\pi$, $\cos\theta=-1$, so $V = -\frac{1}{4\pi\epsilon_0} \frac{p}{r^2}$.
The work done in conservative fields like electrostatic fields is dependent on:
The NCERT text clearly states: 'Work done is independent of the path' (page 48) and 'The work done corresponding to the later will be zero' (page 49). This implies that work done in an electrostatic field depends only on the initial and final positions, as shown in Example 2.1 where 'work done will be path independent'.
For a system of multiple point charges $q_1, q_2, ..., q_n$, the total electric potential at a point P is found using which principle?
The NCERT text states: 'By the superposition principle, the potential V at P due to the system of charges is the algebraic sum of the potentials due to individual charges.' (Page 51).
Consider a point charge Q. If the electrostatic potential at infinity is chosen to be zero, what is the electric potential V at a distance $r$ from this charge Q?
According to the NCERT text (Eq. 2.8), the potential at P due to the charge Q is $V(r) = \frac{1}{4\pi\epsilon_0} \frac{Q}{r}$.
The electric potential due to a dipole is axially symmetric about its dipole moment vector $\vec{p}$. What does this mean?
The NCERT text explains: '(It is, however, axially symmetric about $\vec{p}$. That is, if you rotate the position vector $\vec{r}$ about $\vec{p}$, keeping $\theta$ fixed, the points corresponding to P on the cone so generated will have the same potential as at P.)' (Page 51).
If $\displaystyle \oint_S \vec{E} \cdot d\vec{S} = 0$ over a surface, then:
Zero net flux means inward and outward flux through the surface are equal (by Gauss's law, enclosed charge is zero).
An electric dipole is placed at an angle of $30^\circ$ with an electric field of intensity $2\times 10^5\ \text{N C}^{-1}$. It experiences a torque equal to $4\ \text{N m}$. Calculate the magnitude of charge on the dipole, if the dipole length is $2\ \text{cm}$.
$\tau = qLE\sin\theta\Rightarrow q = \dfrac{4}{0.02 \times 2\times 10^5 \times 0.5} = 2\times 10^{-3}\ \text{C} = 2\ \text{mC}$.
The equivalent capacitance of the system shown in the following circuit is:
The two 3 μF in the parallel branch combine to 6 μF; in series with the leftmost 3 μF: $\dfrac{3\times 6}{3+6} = 2\ \mu\text{F}$.
An electric dipole is placed as shown in the figure.
The electric potential (in $10^2\ \text{V}$) at point $P$ due to the dipole is ($\epsilon_0 =$ permittivity of free space and $\dfrac{1}{4\pi\epsilon_0} = K$):
$P$ on axial line; $V = Kq\!\left(\dfrac{1}{0.02} - \dfrac{1}{0.08}\right) = 37.5\,Kq\ \text{V} = (3/8)\,Kq\ (\times 10^2\ \text{V})$.
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