Question

two particles woth each charge magnitude 2.0×10^-7 c but opposite signs are held 15cm apart.what are the magnitude and direction of the electric field E at tge point midway between charges​

Answer 1

The magnitude of the electric field strength is 6.4 x 10⁵ N/C, directed from positive particle to negative particle.

Step-by-step explanation:

Given;

charge of each particle, Q = 2 x 10⁻⁷ C

distance between the two charges, r = 15 cm = 0.15 m

distance midway between the charges = 0.075 m

The magnitude of the electric field is calculated as;

`E_(net) = E_(+q) + E_(-q)\\\\E_(net) = (kQ)/(r_(1/2)^2) + (kQ)/(r_(1/2)^2)\\\\E_(net) = 2((kQ)/(r_(1/2)^2))\\\\E_(net) = 2 ((9* 10^9 \ * 2* 10^(-7))/(0.075^2) )\\\\E_(net) = 6.4* 10^5 \ N/C`

The direction of the electric field is from positive particle to negative particle.