Question

Two charges Q =2 C and q=1 nC are 1 m apart. If the electric force beween them is 18 N what is the magnitude of the electric field due to Q at q. The format of the answer must be g

Answer 1

Answer:

magnitude of the electric field = 1.8 × $10^{10}$ N/C

Explanation:

given data

charge Q =2 C

charge q = 1 nC  = 1 × $10^{-9}$ C

distance r = 1 m

electric force = 18 N

solution

we know that here electric field due to Q is

E = k × $\frac{Q}{r^2}$   ..............1

here q distance is 1 m

now we  apply here Coulomb’s law and get here electric force

F = k ×$\frac{Q*q}{r^2}$   ..........2

we know constant k  = 8.988 × $10^{9}$ Nm²/C²

so from above both equation we get

electric filed = $\frac{force}{charge}$    .................3

put here value

electric filed = $\frac{18}{1*10^{-9}}$ = 1.8 × $10^{10}$ N/C

Answer 2

Answer:

$E=1.8\times 10^{10}\ N.C^{-1}$

Explanation:

Given:

• one charge, $Q=2\ C$

• another charge, $q=10^{-9}\ C$

• distance between the two charges, $r=1\ m$

• force between the charges, $F=18\ N$

We know from the Coulomb's law:

$F=\frac{1}{4\pi.\epsilon_0} \times \frac{Q.q}{r^2}$ ...........(1)

where:

$\epsilon_0=$ permittivity of free space

Also we have electric field due to Q at q (which is at a distance of 1 m):

$E=\frac{1}{4\pi.\epsilon_0} \times \frac{Q}{r^2}\ [N.C^{-1}]$ ...........(2)

From (1) & (2)

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

$E=\frac{18}{10^{-9}}$

$E=1.8\times 10^{10}\ N.C^{-1}$