Question

The value of the electric field at a distance of 83.9 m from a point charge is 46 N/C and is directed radially in toward the charge. What is the charge? The Coulomb constant is 8.98755 × 109 N · m2 /C 2 . Answer in units of C.

Answer 1

$3.06 \times 10^{-5} C$ is the charge

Explanation:

The electric fields can be defined as the ratio of electric forces per unit charge.

         $E=\frac{F}{q}$ ---------> eq 1

The electric field differs from many quantities such as mass, forces, and speed. The electric field is not a property of a particular object, but a property of a given point in space. All electric charges form an electric field and we can analyse the value (size and direction) of this field at any point in space around the distribution of the charge.

The simplest formula for an electric field is that produced by a point charge:

       $E=\frac{k \times q}{r^{2}}$ ------> eq. 2

Where,

q - Charge

r - Distance of separation from the charge

k = the Coulomb constant (  $8.99 \times 10^{9} \mathrm{Nm}^{2} / \mathrm{C}^{2}$)

If the charge is negative, the field is radially inward; if the charge is positive, it is directed radially outward.

Here, given data:

r = 83.9 m

E = 46 N/C

We need to find q.

Substitute the given values in eq 2, we get

         $46=\frac{8.99 \times 10^{9} \times q}{(83.9)^{2}}$

         $q=\frac{46 \times 83.9 \times 83.9}{8.99 \times 10^{9}}=\frac{323803.66}{8.99 \times 10^{9}}=30618.204 \times 10^{-9}=3.06 \times 10^{4} \times 10^{-9}$

         $q=3.06 \times 10^{-5} \mathrm{C}$