Question

What is the magnitude of a point charge in coulombs whose electric field 68 cm away has the magnitude 2.4 N/C? Units in C

Answer 1

Charge,
`q=1.23* 10^(-10)\ C`

Step-by-step explanation:

It is given that,

Electric field due to a point charge, E = 2.4 N/C

Distance from the point charge, d = 68 cm = 0.68 m

The formula to find the electric field for a point charge at a distance d is given by :

`E=(kq)/(d^2)`

`q=(Ed^2)/(k)`

`q=(2.4* (0.68)^2)/(9* 10^9)`

`q=1.23* 10^(-10)\ C`

So, the magnitude of a point charge is
`1.23* 10^(-10)\ C`. Hence, this is the required solution.

Answer 2

charge will be
`0.123* 10^(-9)C.`

Step-by-step explanation:

We have given that electric field due to point charge is E=2.4 N/C

Distance from point charge R = 68 cm = 0.68 m

We know that electric field due to point charge is given as
`E=(1)/(4\pi \epsilon _0)(Q)/(R^2)=(KQ)/(R^2)`

So
`2.4=(9* 10^9Q)/(0.68^2)`

`Q=0.123* 10^(-9)C`

So charge will be
`0.123* 10^(-9)C.`