Question

1.) Two charges, q₁ = -16.5 μC and q₂ = 13.5 μC , are located at (x,y) = (12.4, 21.2) cm and (19.6, 24.0) cm respectively. Find the electrostatic force between these two changes (take an attractive force to be negative)

2.)Two charges, q₁ =-16.1 μC and q₂ = 11.9 μC , are located along a straight line at x = 19.5 cm and 57.5 cm respectively. What is the electric field at a point P located at x = 35.1 cm?[Consider a field directed to the right as being positive]

3.) Two plates, separated by a distance of 15.0 cm, have a potential difference of 3.40×10³ between them, with the positive plate on the left. Calculate the electric field between the plate, taking as positive a field that points to the right.

4.) Two charges, q₁ = 34 μC and q₂ = 40 μC, are located along a straight line at x = 34 cm and 64 cm respectively. Where along the x-axis should a third charge of q₃ = -11 μC be located such that the net force on this charge is zero.

Answer 1

1.

Given  

q1 = -16.5 μC and q2 = 13.5 μC , are located at (x,y) = (12.4, 21.2) cm and (19.6, 24.0) cm respectively

From Coulomb's law

F = kq1*q2/r^2

F = 9*10^9((-16.5*10^-6)(13.5*10^-6)/((19.6-12.4)^2+(24-21.2)^2)) N = -0.033591655496 N

F = -33.591655496*10^-3 N

attractive force  

2.Two charges, q1 =-16.1 μC and q2 = 11.9 μC , are located along a straight line at x = 19.5 cm and 57.5 cm respectively.

the point P located is x = 35.1 cm

the electric field is E = E1+E2

E1 = kq1/x1, E2 = kq2/x2

E1 = (9*10^9*16.1*10^-6)/((35.1-19.5)(10^-2)) N/C = 928846.1538 N/C

E2 = (9*10^9*11.9*10^-6)/((57.5-35.1)(10^-2)) N/C = 478125 N/C

we knwo that the field lines are radially out ward for +ve charges andradially out ward for -ve charges  

so the net field at point P is  

E = E1+E2 = 928846.1538 +478125 N/C = 1406971.1538 N/C

by symmetry the direction is to the left

3.

we knwo that the electric field , Potential and distance relation i s

V = E*d ==> E = V/d = 3.40*10^3/0.15 V/m = 22666.667 V/m

4.

Given charges q1 = 34 μC and q2 = 40 μC, are located along a straight line at x = 34 cm and 64 cm respectively.

let the third charge q3 = -11 μC be located near the lower charge say at x distance from q1 then it would be at a distance

(30-x) from q2

now equating the forces by q1,q2 on q3

F13 = F23

kq1q3/x^2 = k*q2q3/(0.03-x)^2

q1/x^2 = q2/(0.03-x)^2

34*10^-6/(x^2) = (40*10^-6)/(0.03-x)^2

solving for x

x = 0.01438 m = 1.438 cm from q1