Question

The electric field has a magnitude of 3.0 N/C at a distance of 60 cm from a point charge. What is the
charge?​

Answer 1

Answer:

1.2x 10^-9C

Explanation:

ELECTRIC FIELD AT A POINT IS (9X 10⁹X Q)/(0.6)²

Answer 2

The value of the charge will be 120 × $10^{-12}$ C.

Coulomb's Law: The electric field between two point charges will be directly proportional to their magnitude.

Step-by-step explanation:

Given: Electric Field (E) = 3.0 N/C

           Distance(d) = 60cm or 0.60m

To Find: Electric charge(q)

Formula used:        E = $(\frac{Kq }{r^{2} } )$       ( K = proportionality constant)

                                3 = $(\frac{9\times10^{9}\times q }{(0.6)^{2} } )$

solving this, we get charge(q) = 120 × $10^{-12}$ Coulomb.

Therefore, the magnitude of charge will be 120 pC.