Point charges of 28.0 µc and 35.0 µc are placed 0.250 m apart. (a) at what point (in m) along the line between them is the electric field zero?m away from the 28.0 µc charge (b) what (in n/c) is the electric field halfway between them? (enter the magnitude.)

5 ZERO FIELD POINT BETWEEN TWO CHARGES

Draw a horizontal arrow for an x axis.

28.0 charge goes at x=0.

35.0 charge goes at x=0.25

Zero field ...

28/x^2 = 35/(0.25-x)^2

Quadratic in x.

Solve quad.

poss answer

(root 5-2)/2

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