Derive an equation for electric potential V(z) on z axis

In summary, the conversation revolves around a homework problem involving creating an equation for V(z) using the V=kQ/r equation and a diagram with four negative charges on a positive axis. The relevant equations are E=F/q, F=kQq/r^2, V=kQq/r, U=V*q, KE=0.5m*velocity^2, and k=1/4*pi*Enod. The student has completed parts (a) and (b) but is struggling with making the equation for V(z) negative and is seeking help for parts (c), (d), and (f). They have derived an equation for V(z) as V(z)=Q/(4*pi*Enod)*r where
  • #1
jackMybrain@ru
6
1

Homework Statement


Hi, I am stuck with this homework. I have been asked to make an equation out of a diagram for V(z) using V=kQ/r equation, where z is a positive axis centred with four negative charges. Here is the diagram.
9f12337b48.png


2. Relevant equation
E=F/q
F of point charge: F=kQq/r^2
E=kQ/r^2
V=kQq/r ~U=V*q & KE=0.5m*velocity^2

r=distant of the point from the point charge q
k=1/4*pi*[E][/0]

The Attempt at a Solution


I have already completed the (a) and (b) parts. However, for part (b) I could only make V(z)=Q/(4*pi*Enod)*r where r = 0.5*sqrt([d^2+(2Z-d)^2]) from the r12. I found r by taking all point charges distant together by saying r12 of the first 2 charges are equal and r34 of the 3rd and 4th is same. Therefore, Vtotal=KQ*[(2/r12)+(2/r34)]. But how do I make it as V(z)=-Q/pi*Enod... because currently I am getting V(z)=-Q/8*pi*Enod(sqrt([d^2+(2Z-d)^2])).
Am I doing something wrong. Thanks for any insight or solution for the other parts would be very helpful.
And how do I do Q 1. (c),(d)& (f)??
 
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  • #2
jackMybrain@ru said:

Homework Statement


Hi, I am stuck with this homework. I have been asked to make an equation out of a diagram for V(z) using V=kQ/r equation, where z point start in the center with four negative charges. Here is the diagram for the aid.
9f12337b48.png


2. Relevant equation
E=F/q
Force of point charges by Coulumbs law: F=kQq/r^2
E=kQ/r^2
V=kQq/r ~U=V*q & KE=0.5*mass*velocity^2

r=distant of the point from the point charge q
k=1/4*pi*Enod

The Attempt at a Solution


I have already completed the (a) and (b) parts.
However, for part (b) I don't understand how to make a negative result for V(z)??
As normal equation goes, I have derived only V(z)=Q/(4*pi*Enod)*r where r = 0.5*sqrt([d^2+(2Z-d)^2]) (from the r12).
When I put everything together my found equation: V(z)=-Q/8*pi*Enod(sqrt([d^2+(2Z-d)^2])).
But how do I make it as V(z)=-Q/pi*Enod.(...) because both distance and z axis are positive.

**I found my d or r by taking all point charges distant from z together. For instance accumulating r12 of q1 and 12 are equal apart from z on z axis and r34 of the 3rd and 4th are higher than them vice versa. Therefore, (r) total=(2/r12)+(2/r34). So, Vtotal=KQ/[(2/r12)+(2/r34)].

Am I doing something wrong. I am only trying to understand physics. Thanks for any insight or solution for the other parts would be very helpful.
And how do I do Q 1. (c),(d)& (f)??

This is an easier version of my question. Please, help! I don't know why nobody is familiar with V(z) equation or how to solve them. Because there are not many references or videos to be found by online searching. Thanks again..
 
  • #3
Show your derivation for r12 and r34 in detail.
 

Related to Derive an equation for electric potential V(z) on z axis

1. What is the equation for electric potential V(z) on the z axis?

The equation for electric potential V(z) on the z axis is given by V(z) = kQ/z, where k is the Coulomb's constant, Q is the point charge, and z is the distance from the point charge.

2. How is the electric potential V(z) on the z axis related to the electric field?

The electric potential V(z) on the z axis is related to the electric field by the equation E(z) = -dV(z)/dz, where E(z) is the electric field and dV(z)/dz is the derivative of the electric potential with respect to z.

3. Can the equation for electric potential V(z) on the z axis be used for any type of charge distribution?

Yes, the equation for electric potential V(z) on the z axis can be used for any type of charge distribution as long as the distance z is measured from the center of the charge distribution.

4. How does the electric potential V(z) on the z axis change with distance?

The electric potential V(z) on the z axis decreases with distance. As the distance z increases, the electric potential decreases and approaches zero at an infinite distance.

5. How is the electric potential V(z) on the z axis affected by the presence of multiple point charges?

The electric potential V(z) on the z axis is the sum of the individual electric potentials from each point charge. Therefore, the presence of multiple point charges will affect the overall electric potential on the z axis.

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