NEET UG PHYSICS MCQ: ELECTRIC FIELDS AND CHARGES

1 Hard

1 of 1

Application +4 | -1

Two point charges are kept in air with certain separation between them. The force between them is

F_1

, if half of the space between the charges is filled with a medium of dielectric constant 4 and the force between them is

F_2

, if

\frac{1}{3}

rd of the space between the charges is filled with dielectric of dielectric constant 9. Then

F_1 / F_2

View answer and explanation

Correct option: A

\frac{100}{81}

Option A (100/81) is correct because it accurately calculates the ratio of forces with and without the dielectric, considering the effective distance change.

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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