1.) Two charges, q 1 = -16.5 C and q 2 = 13.5 C , are located at (x,y) = (12.4,
ID: 1580549 • Letter: 1
Question
1.) Two charges, q1 = -16.5 C and q2 = 13.5 C , 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 C and q2 = 11.9 C , 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 C and q2 = 40 C, 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 C be located such that the net force on this charge is zero.
Explanation / Answer
1.
Given
q1 = -16.5 C and q2 = 13.5 C , 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 C and q2 = 11.9 C , 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 C and q2 = 40 C, are located along a straight line at x = 34 cm and 64 cm respectively.
let the third charge q3 = -11 C 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
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