1
KCET 2023
MCQ (Single Correct Answer)
+1
-0

In the situation shown in the diagram, magnitude, if $$q < < |Q|$$ and $$r >>>a$$. The net force on the free charge $$-q$$ and net torque on it about $$O$$ at the instant shown are respectively.

($$p=2 a Q$$ is the dipole moment)

KCET 2023 Physics - Electrostatics Question 10 English

A
$$\frac{1}{4 \pi \varepsilon_0} \frac{p q}{r^3} \hat{\mathbf{i}},-\frac{1}{4 \pi \varepsilon_0} \frac{p q}{r^2} \hat{\mathbf{k}}$$
B
$$\frac{1}{4 \pi \varepsilon_0} \frac{p q}{r^2} \hat{\mathbf{k}}, \frac{1}{4 \pi \varepsilon_0} \frac{p q}{r^3} \hat{\mathbf{i}}$$
C
$$-\frac{1}{4 \pi \varepsilon_0} \frac{p q}{r^2} \hat{\mathbf{k}},-\frac{1}{4 \pi \varepsilon_0} \frac{p q}{r^3} \hat{\mathbf{i}}$$
D
$$\frac{1}{4 \pi \varepsilon_0} \frac{p q}{r^3} \hat{\mathbf{i}},+\frac{1}{4 \pi \varepsilon_0} \frac{p q}{r^2} \hat{\mathbf{k}}$$
2
KCET 2023
MCQ (Single Correct Answer)
+1
-0

A uniform electric field vector $$\mathbf{E}$$ exists along horizontal direction as shown. The electric potential at $$A$$ is $$V_A$$. A small point charge $$q$$ is slowly taken from $$A$$ to $$B$$ along the curved path as shown. The potential energy of the charge when it is at point $$B$$ is

KCET 2023 Physics - Electrostatics Question 11 English

A
$$q\left[V_A-E x\right]$$
B
$$q\left[V_A+E x\right]$$
C
$$q\left[E x-V_A\right]$$
D
$$q E x$$
3
KCET 2023
MCQ (Single Correct Answer)
+1
-0

A cubical Gaussian surface has side of length $$a=10 \mathrm{~cm}$$. Electric field lines are parallel to $$X$$-axis as shown in figure. The magnitudes of electric fields through surfaces $$A B C D$$ and $$E F G H$$ are $$6 ~\mathrm{kNC}^{-1}$$ and $$9 \mathrm{~kNC}^{-1}$$ respectively. Then, the total charge enclosed by the cube is

[Take, $$\varepsilon_0=9 \times 10^{-12} \mathrm{~Fm}^{-1}$$ ]

KCET 2023 Physics - Electrostatics Question 13 English

A
$$-0.27$$ nC
B
1.35 nC
C
$$-1.35$$ nC
D
0.27 nC
4
KCET 2023
MCQ (Single Correct Answer)
+1
-0

Electric field at a distance $$r$$ from an infinitely long uniformly charged straight conductor, having linear charge density $$\lambda$$ is $$E_1$$. Another uniformly charged conductor having same linear charge density $$\lambda$$ is bent into a semicircle of radius $$r$$. The electric field at its centre is $$E_2$$. Then

A
$$E_2=\pi r E_1$$
B
$$E_2=\frac{E_1}{r}$$
C
$$E_1=E_2$$
D
$$E_1=\pi r E_2$$
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