If A = 3.0 Mm B = 4.0 Mm Q 1 = 60 Nc Q 2 = 80 Nc And Q = 24 Nc In The Figure What Is The

In the figure, if Q = 30 C, q = 5.0 C, and d = 30 cm, what is the magnitude of the electrostatic force on q?

If Q = 30 μC and q = 5 μC

Force due to Q on q = F1
F1 = kQq/d^2
= 9*10^9*30*5*10^-12 / (0.3)^2
= 15 N

Force due to 2Q on q = F2
F2 = k*2Q*q/(2d)^2
= kQq/(2d^2)
= F1/2
= 15/2
= 7.5N

Resultant force acting on q is ΔF
ΔF = 15 - 7.5 = 7.5 N

If a = 60 cm, b = 80 cm, Q = -4.0 nC, and q = 1.5 nC, what is the magnitude of the electric field at point?

If a = 60 cm, b = 80 cm, Q = -4.0 nC, and q = 1.5 nC, what is the magnitude of the electric field at point P?

picture:
http://hccs.blackboard.com/webct/RelativeResourceManager/Template/RspQ-Chapter%2023%20Electric%20Fields/mc022-1.jpg

a. 68 N/C
b. 72 N/C
c. 77 N/C
d. 82 N/C
e. 120 N/C

HELP i need step by step on this one.

The two charges in the figure below are separated by d = 2.50 cm. (Let q1 = −19.5 nC and q2 = 30.0 nC.)?

a) To find the electric potential at point A, we must find the potential due to each charge individually, then just add them. So, using Coulomb's law for potential we have

P(q1) = K q1 / r J
where K = Coulomb's constant = 9 * 10^9 ; r = 0.025 m

P(q2) = K q2 / r J

P(A) = P(q1) + P(q2) = (K/r) ( q1 + q2)

P(A) = 9 * 10^9 [(-19.5 * 10^-9) + (30.0 ^ 10^-9)] / 0.025

P(A) = 9 * 10^9 ( 10.5 * 10^-9) / 0.025

P(A) = 3.78 kJ

b) We have an isoceles triangle, therefore the base angles are equal. The angle at A is (180 - 120) = 60 degrees. So we now know that it is an equilateral triangle. So the distance between the charges must be 2.5 cm or 0.025 m.
Therefore, the half way point is 0.025/2 = 0.0125 m

P(B) = (K/r) (q1 + q2)
where, this time, r = 0.0125 m

P(B) = (9 * 10^9 / 0.0125) ( 10.5 * 10^-9) = 7.56 kJ

If a = 30 cm, b = 20 cm, q = +2.0 nC, and Q = –3.0 nC?

The potential due to a point charge, at a distance "r" from the charge is;

V(r) = kq/r

If you have more then one point charge you add , algebraically, the potential due to each. You use the correct sign of the charge. So

V(A) = kq/a + kQ/(a+b)

=k[2x10^-9)/(.3) - (3x10^-9)/(.5)]
= k(.67x10^-9)
= (9x10^9)(.67x10^-9) = 6 V

V(B) = kq/(a+b) + kQ/a

=(9x10^9)[(2x10^-9)/(.5) - (3x10^-9)/(.3)]
= -54 V

SO,

V(A) - V(B) = 6 - (-54) = 60 V

Physics: If Q = 16 nC, a = 3.0 m, and b = 4.0 m, what is the magnitude of the electric field at point P?

Force P = Vector sum of force along 3m & force along 4m = √[f₃²+f₄²];
f₃² = Q²/r² Q²/9,
f₄² = Q²/r² Q²/16,
P =Q²√[1/9 + 1/16] = [Q²/(3X4)]√[16+8] = [Q²/(3X4)]√[25] = (5/12) Q².

If Q=16nC, α=3.0m , and b=4.0 m what is the magnitude of the electric field at point P?

Figure Drawing: (rectangle with 3 charges, Q, at 3 of the 4 corners. The remaining 4th corner is marked point P. The length of the rectangle is labeled b, the width a. One corner is labeled to be a 90 degree angle, but obviously all corners in a rectangle are 90 degrees.)

I don't even know where to start. My mind automatically sees the pythagoraen theorem but I assume integration would be the most appropriate approach to the problem. i'm lost.