NEET UG PHYSICS MCQ: ELECTRIC FIELDS AND CHARGES

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Application +4 | -1

Force between two identical charges placed at a distance of

r

in vacuum is

F

. Now a slab of dielectric of dielectric constant 4 is inserted between these two charges. If the thickness of the slab is

r/2,

then the force between the charges will become

View answer and explanation

Correct option: D

\frac{F}{4}

Option D (F/4) is correct because the introduction of a dielectric slab effectively reduces the electric field and hence the force between the charges.

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The ratio of magnitude of electric displacement (D) and electric field (E) is Gauss’s Law is valid for In hydrogen like system the ratio of coulombian force and gravitational force between an electron and a proton is in the order of: Statement-1:

\overrightarrow{{E}}

in outside velocity of a conductor depends only on the local charge density

\alpha

.
Statement-2:

\overrightarrow{{E}}

in outside vicinity of a conductor is given by

\frac{\sigma}{\varepsilon_0}

. Two charges each of 1 coulomb are at a distance 1

km

apart the force between them is

P \times 10^3

then value of

{P}

is Electric field due to point charge depends on distance

r