1
NEET 2024 (Re-Examination)
+4
-1

The value of electric potential at a distance of $$9 \mathrm{~cm}$$ from the point charge $$4 \times 10^{-7} \mathrm{C}$$ is [Given $$\frac{1}{4 \pi \varepsilon_0}=9 \times 10^9 \mathrm{~N} \mathrm{~m}^2 \mathrm{C}^{-2}$$] :

A
$$4 \times 10^2 \mathrm{~V}$$
B
$$44.4 \mathrm{~V}$$
C
$$4.4 \times 10^5 \mathrm{~V}$$
D
$$4 \times 10^4 \mathrm{~V}$$
2
NEET 2024
+4
-1

A thin spherical shell is charged by some source. The potential difference between the two points $$C$$ and $$P$$ (in V) shown in the figure is:

(Take $$\frac{1}{4 \pi \varepsilon_0}=9 \times 10^9$$ SI units)

A
$$3 \times 10^5$$
B
$$1 \times 10^5$$
C
$$0.5 \times 10^5$$
D
Zero
3
NEET 2024
+4
-1

Given below are two statements: one is labelled as Assertion A and the other is labelled as Reason R.

Assertion A: The potential (V) at any axial point, at $$2 \mathrm{~m}$$ distance $$(r)$$ from the centre of the dipole of dipole moment vector $$\vec{P}$$ of magnitude, $$4 \times 10^{-6} \mathrm{C} \mathrm{m}$$, is $$\pm 9 \times 10^3 \mathrm{~V}$$.

(Take $$\frac{1}{4 \pi \epsilon_0}=9 \times 10^9 \mathrm{SI}$$ units)

Reason R: $$V= \pm \frac{2 P}{4 \pi \epsilon_0 r^2}$$, where $$r$$ is the distance of any axial point, situated at $$2 \mathrm{~m}$$ from the centre of the dipole.

In the light of the above statements, choose the correct answer from the options given below:

A
Both A and R are true and R is the correct explanation of A.
B
Both A and R are true and R is NOT the correct explanation of A.
C
A is true but R is false.
D
A is false but R is true.
4
NEET 2023 Manipur
+4
-1

According to Gauss law of electrostatics, electric flux through a closed surface depends on :

A
the area of the surface
B
the quantity of charges enclosed by the surface
C
the shape of the surface
D
the volume enclosed by the surface
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