What is the Electric Field and Spring Constant for Two Charges on a Spring?

[jdmarquardt]

Homework Statement

Two charges Qa = 3 µC and Qb = -2 µC are placed on the x-axis with a separation of a = 25 cm.

(a) Find the net electric field at point P, a distance d = 7 cm to the left of charge Qa.
(b) Find the force on Qb due to Qa .
The charges Qa and Qb are now attached to the ends of a spring whose unstretched length is s0 = 25 cm. With the charges attached, the spring compresses to an equilibrium length s1 = 10 cm.

(c) Calculate the spring constant ks of the spring.

Homework Equations

E=F/q

 

[cepheid]

Welcome to PF,

What have you done so far on this problem?

 

[jdmarquardt]

I am confused from the start. I am unsure on how to find the EF for point P without its Charge, and unsure how to approach it.
Also I am horrible when working to find the spring constant. Any suggestions on how to start, besides a FBD.
Thank you

 

[cepheid]

I am confused from the start. I am unsure on how to find the EF for point P without its Charge, and unsure how to approach it.

The word in bold above indicates that you have a misunderstanding. There does not need to be a charge at point P in order for there to be an electric field there. The electric field you are looking for is the field due to the other charges. The electric fields of Q_a and Q_b extend throughout space, including to point P. Basically you can think of the electric field as "map" of how an electric charge influences its surroundings. Does that make sense?

You're missing an equation for the electric field of a charge q that tells you how strong it is at a distance r away from that charge. You need to find this equation so that you can use it to compute the strength of charge a and charge b's field at point P.

Also I am horrible when working to find the spring constant. Any suggestions on how to start, besides a FBD.
Thank you

Since the charges have opposite signs, they are attracted towards each other. However, since they are at opposite ends of the spring, the spring gets compressed as the charges move towards each other. However, as you know, the spring fights back with a restoring force that gets bigger the more you compress the spring. When the outward force from the spring is equal to the inward force from the electric charges, the compression will stop. So, the problem is telling you that, when the spring is compressed down to s = 10 cm, the spring force and the electric force are equal to each other. That means you know the spring force and you know the amount of compression. So how would you solve for k?

 

[Nickie]

F=kq1q2/r^2
F=-kx

The net electric field at point P is the vector sum of the electric fields due to each charge. Using the equation E=F/q, we can calculate the electric field due to each charge separately.

(a) Electric field due to Qa:
Ea = kQa/d^2 = (9x10^9 Nm^2/C^2)(3x10^-6 C)/(0.07 m)^2 = 6.12x10^6 N/C

Electric field due to Qb:
Eb = kQb/(a-d)^2 = (9x10^9 Nm^2/C^2)(-2x10^-6 C)/(0.18 m)^2 = -1.23x10^6 N/C

Net electric field at P:
E = Ea + Eb = (6.12x10^6 N/C) + (-1.23x10^6 N/C) = 4.89x10^6 N/C

(b) The force on Qb due to Qa can be calculated using the equation F=kq1q2/r^2.
F = (9x10^9 Nm^2/C^2)(3x10^-6 C)(-2x10^-6 C)/(0.18 m)^2 = -1.23x10^-2 N

(c) The force exerted by the spring is equal and opposite to the force calculated in part (b). Therefore, we can use the equation F=-kx to solve for the spring constant.
-kx = -1.23x10^-2 N
-k(0.1 m) = -1.23x10^-2 N
k = 1.23x10^-1 N/m

Therefore, the spring constant ks is 1.23x10^-1 N/m.