KCET 2018 | Electrostatics Question 43 | Physics

The force of repulsion between two identical positive charges when kept with a separation $r$ in air is $F$. Half the gap between the two charges is filled by a dielectric slab of dielectric constant $=4$. Then, the new force of repulsion between those two charges becomes

$\frac{F}{3}$

$\frac{F}{2}$

$\frac{F}{4}$

$\frac{4 F}{9}$

KCET 2018

MCQ (Single Correct Answer)

-0

The work done to move a charge on an equipotential surface is

infinity

less than 1

greater than 1

zero

KCET 2017

MCQ (Single Correct Answer)

-0

Two point charges $A=+3 \mathrm{nC}$ and $B=+1 \mathrm{nC}$ are placed 5 cm apart in air. The work done to move charge $B$ towards $A$ by 1 cm is

$2.0 \times 10^{-7} \mathrm{~J}$

$2.7 \times 10^{-7} \mathrm{~J}$

$12.1 \times 10^{-7} \mathrm{~J}$

$1.35 \times 10^{-7} \mathrm{~J}$

KCET 2017

MCQ (Single Correct Answer)

-0

$4 \times 10^{10}$ electrons are removed from a neutral metal sphere of diameter 20 cm placed in air. The magnitude of the electric field (in $\mathrm{NC}^{-1}$ ) at a distance of 20 cm from its centre is

640

5760

Zero

1440

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