Question

Please answer all parts. Thank you!

PART A)

A point charge gives rise to an electric field of 3 N/C at a distance of 7 m. If the distance remains the same and the charge is increased by a factor of 5, then what will be the new electric field? N/C Submit Answer Tries 0/2

PART B &C)

Answer 1

We know that electric field procedure by point charge is given by E^vector = kg /r^2 Now E^vector = 3 N/c and r = 7 m then 3 N/c = kg/97m)^2 Now Q becomes 5 Q then E^vector = k (SQ)/(7 m)^2 = 5 k Q/(7 m)^2 5 times 3 N/C therefore Epsilon^vector_new = 15 N/c Answer = 15 Now E^vector = k Q/R^2 Given Q = negative 4 Q and R = IR then E^vector = k (negative 4 Q)/(IR)^2 = negative 4 k Q/R^2 therefore Vertical-bar Epsilon^vector Vertical-bar = 4 KQ/R^2 Answer = 4 Now electric force is given by F = k q_1 q_2/r^2 Given q_1 = negative 4 Q and q_g = negative 2 Q and r = 1 R then F = k (negative Q) (2 a)