Question

if a dielectric material of dielectric constant €r, is completely filled in the
space around the point changes shown in figure
then the net electrostatic
force on q1 is​

Answer 1

Explanation:

The distance between q1 and q2 is r. The dielectric constant is represented as εr.

Electrostatic force $=k\frac{q_1 q_2}{r^2}$

We can write f=1/4π∈($\frac{q_1 q_2)}{r^2}$

When $2$ charges have emerged in dielectric force, it reduces by εr

New force = f/εr i.e previous force value divided by εr

$f_(new) = f r$

f(new) =1/4π∈r $\frac{q_1 q_2}{r^2}$

Answer 2

Answer:

The force on charge q1 is $F=\frac{1}{4\pi \epsilon \epsilon _{r}} *\frac{q_{1}q_{2}}{r^2}$

Explanation:

• As we know that the force on a charge q is
  $F=\frac{KqQ}{r^2}$
• Now we kalso know that the value of
  $K=\frac{1}{4\pi \epsilon}$
• As the dielectric material of dielectric constant $\epsilon_{r}$ is filled so the value of new force is $F_{new} = \frac{F}{\epsilon_{r}}$
• Hence the changed force is
  $F=\frac{1}{4\pi \epsilon \epsilon _{r}} *\frac{q_{1}q_{2}}{r^2}$
  #SPJ2