Point charge q1 and q2 both of 22nc are separated by a distance of 58cm along horizontal axis point a is located 40cm from q1 and 18 cm from q2. what is the magnitude and direction of the electric field at point a

E = 4.01×10^6 N/C

direction is toward q2

Explanation:

q1 = q2 = 12micro-C = 12×10^-6C

r1 = 18cm = 0.18m

r2 = 40cm = 0.40m

k = 9×10^9

E1 = kq1/(r1)²

E1 = 3.33×10^6 N/C

E2 = kq2/(r2)²

E2 = 6.75×10^5 N/C

Enet = E1 + E2

= 3.33×10^6 + 6.75×10^5

= 4.01×10^6 N/C

direction is toward q2

d.7.5cm is your answer

initial speed when it is dropped is 150 ft/s

now at the same moment the acceleration due to gravity is perpendicular to its velocity

so as we know that

now we have centripetal acceleration a = 32 ft/s^2 as its perpendicular to velocity

while tangential acceleration at the same moment will be zero

so

now radius of curvature is given as

so the radius of curvature just after it is dropped is 703.125 ft