Question

The electric field at point P due to a point charge Q a distance R away has magnitude E. In order to double the magnitude of the field at P, you could?

Answer 1

The electric field at point P due to a point charge Q at a distance R away has magnitude E. In order to double the magnitude of the field at P, you could double the charge Q or keeping the same charge you can introduce a medium of permittivity $\epsilon_2$ as detailed below.

An electric field is the physical field that surrounds electrically charged particles and exerts force on all other charged particles in the field, either attracting or repelling them. Electric field is defined as the electric force per unit charge i.e.

$E = \frac{F}{q}$

Electric field due to a point charge q at a point P which is at distance d is defined as,

$E = \frac{kq}{d^2}$ ;      $k = \frac{1}{4\pi \epsilon}$   where $\epsilon$ is the permittivity of medium.

• If the distance ($d$) is constant, for a given medium $k$, the electric field is directly proportional to charge $q$ i.e $E \propto q$.  

       If the electric field strength $E_2$ is twice  $E_1$ then, $\frac{E_2}{E_1} = \frac{q_2}{q_1}$

         $\frac{2}{1} = \frac{q_2}{q_1} \implies q_2 ={2q_1}$

         or $\boxed{q_2 = 2Q}$

        Doubling the charge of the particle will double the electric field at point P which is at distance R.

• If the distance ($d$)is constant, for a given charge $Q$, the electric field is inversely proportional to the permittivity of the medium i.e $E \propto\frac{1}{\epsilon}$.

        If the electric field strength $E_2$ is twice  $E_1$ then,

                $\frac{E_2}{E_1} = \frac{\epsilon_1}{\epsilon_1}$

                       $\frac{2}{1} = \frac{\epsilon_1}{\epsilon_2} \implies \boxed{\epsilon_2 =\sqrt{\frac{\epsilon_1}{2}}}$

We can introduce a medium of permittivity $\epsilon_2$.

For more details on electric field, check:

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