Two questions - Electric Potential

[bcjochim07]

Homework Statement

#1

A -10 nC charge and a +20 nC charge are 15cm apart on the x-axis. What is the electric potential at the spot where electric field = 0?

So called the 20nC charge q1 & the -10 nC charge q2.
I also defined r1 as distance from the 20nC charge, so r2 must then = .15 + q1

so Kq1/(r1)^2 = Kq2/(r1 +.15)^2 but when I put in the values and solve the resulting quadratic, I get a negative radius. If I switch how I assigned the charges, however, I get the correct answer. Why must my r1 be the distance from the -10 nC charge?


#2

Two 10 cm diameter electrodes are .50cm apart to form a parallel plate capacitor. The electrodes are attached to a 15V battery. Insulating handles are used to pull the electrodes away from each other until they are 1 cm apart. The electrodes remain connected to the battery during this process. What is the electric field after this process?

When they aren't pulled away deltaV = (.005m)E E= 3000V/m

I'm thinking that E should be constant because E = surface charge density / permittivity constant, but when they are pulled apart I find that:

V=Ed (15 V ) = E (.01) E = 1500 V/m. How can this change?

Homework Equations

The Attempt at a Solution

 

[Defennder]

For #1:

Draw out the x-axis containing the 2 charges. Let q1 be the one on the left and the other to the right of it. Now from first glance it looks like there are 2 regions one can find the E-field to be zero, namely the space either to the left or right of both charges (excluding the middle).

Suppose, as you did, we picked the space on the left. Where is the maximum electrostatic field strength which can be exerted by q2 to the left? Clearly that would be where q1 resides. Yet at the point, or anywhere on the left close to it, the electric field due to q1 would be much stronger and would more than cancel out the E-field due to the other charge, since the q1 has a larger charge magnitude. Evaluating the E-field at any point further left would decrease E-field due to q1, but it would also decrease the E-field due to q2, and since q2 is smaller in magnitude and further away, the E-field due to q2 would decrease at a rate greater than that due to q1.

Therefore it can't be that left region. So it has to be on the other side of the 2 charges.

#2:

I'm thinking that E should be constant because E = surface charge density / permittivity constant, but when they are pulled apart I find that:

V=Ed (15 V ) = E (.01) E = 1500 V/m. How can this change?

That is true only for an infinite plane of charge (which isn't connected to a battery)

 

[baba89]

#1: The distance between the two charges should not affect the direction of the electric field. The electric field always points from positive to negative charges. Therefore, the distance from either charge should not change the direction of the electric field. However, the magnitude of the electric field will change depending on the distance from the charges.

In this case, it seems that there may be an error in the calculations or in the assignment of the charges. It is important to double check the values and equations used to ensure accuracy.

#2: The electric field is not constant in this case because the distance between the charges is changing. As the plates are pulled apart, the distance between them increases, causing the electric field to decrease. This is because the electric field is inversely proportional to the distance between the plates.

The equation V = Ed is valid only when the electric field is constant. In this case, as the distance between the plates changes, the electric field also changes, so this equation cannot be used.

To calculate the electric field in this case, the equation E = V/d should be used, where V is the potential difference (15V) and d is the distance between the plates (.01m). This will give an accurate value for the electric field after the plates have been pulled apart.