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 23 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 24 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 26 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
$$E_1=\pi r E_2$$
KCET Subjects
EXAM MAP
Medical
NEETAIIMS
Graduate Aptitude Test in Engineering
GATE CSEGATE ECEGATE EEGATE MEGATE CEGATE PIGATE IN
Civil Services
UPSC Civil Service
Defence
NDA
Staff Selection Commission
SSC CGL Tier I
CBSE
Class 12