Question

The electric field at a point 2.8 cm from a small object points toward the object with a strength of 180,000 N/C. What is the object's charge q? ( k = 1/4πε 0 = 8.99 × 10 9 N ∙ m 2/C 2)

Answer 1

Charge,
`Q=1.56* 10^(-8)\ C`

Step-by-step explanation:

It is given that,

Electric field strength, E = 180000 N/C

Distance from a small object, r = 2.8 cm = 0.028 m

Electric field at a point is given by :

`E=(kQ)/(r^2)`

Q is the charge on an object

`Q=(Er^2)/(k)`

`Q=(180000\ N/C* (0.028\ m)^2)/(9* 10^9\ Nm^2/C^2)`

`Q=1.56* 10^(-8)\ C`

So, the charge on the object is
`1.56* 10^(-8)\ C`. Hence, this is the required solution.