1
KCET 2023
+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

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
$$q E x$$
2
KCET 2023
+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, $$\varepsilon_0=9 \times 10^{-12} \mathrm{~Fm}^{-1}$$ ]

A
$$-0.27$$ nC
B
1.35 nC
C
$$-1.35$$ nC
D
0.27 nC
3
KCET 2023
+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
$$E_1=\pi r E_2$$
4
KCET 2022
+1
-0

A tiny spherical oil drop carrying a net charge $$q$$ is balanced in still air, with a vertical uniform electric field of strength $$\frac{81}{7} \pi \times 10^5 \mathrm{~V} / \mathrm{m}$$. When the field is switched OFF, the drop is observed to fall with terminal velocity $$2 \times 10^{-3} \mathrm{~ms}^{-1}$$. Here $$g=9.8 \mathrm{~m} / \mathrm{s}^2$$, viscosity of air is $$1.8 \times 10^{-5} \mathrm{Ns} / \mathrm{m}^2$$ and density of oil is $$900 \mathrm{~kg} \mathrm{~m}^{-3}$$. The magnitude of $$q$$ is

A
$$8 \times 10^{-19} \mathrm{C}$$
B
$$1.6 \times 10^{-19} \mathrm{C}$$
C
$$3.2 \times 10^{-19} \mathrm{C}$$
D
$$0.8 \times 10^{-19} \mathrm{C}$$
EXAM MAP
Medical
NEET