# Electrostatic - electric potential due to a point charge

• polibuda
In summary: Given that we have ##E = 10^4 \frac{V}{m}## at ##r = 0.05 \text{ m}## we can solve for ##Q## (Do this part on your own and tell me what you get)Plug your answer for ##Q## into the following and tell me what you get.##V \left( r = 0.05 \text{ m} \right) = \frac{Q}{4 \pi \epsilon_0 \left( 0.05 \right)}##taking your previous answer for V,then solve the equations##V \pm 100 = \frac{Q}{4 \
polibuda
Homework Statement
The equipotential surface passes through a point with field intensity electric 10 kV / m at a distance from a point charge generating a field of
r1 = 5 cm. At what distance from the field generating charge it belongs
carry out the second equipotential surface to make the potential difference
between these surfaces was equal to 100 V.
Relevant Equations
E=V/d
Could somebody check my solution? I want to know is it correct.

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BvU said:
I'm sorry. Now is it ok?

I think this is a badly worded question. There can be a potential difference between two surfaces that is positive or negative which this problem doesn't specify. That will dictate whether the second surface is closer or further away from the point charge than the first surface. Basically what I'm saying is that without further specification of higher or lower potential we have two possible answers for how far away from the point charge (the one generating the field) the second surface is located.

I will try this problem out for myself and see what I get in about an hour or so using both higher and lower potential (assuming I understand what the problem is saying).

polibuda
polibuda said:
I'm sorry. Now is it ok?
No. It's illegible:
polibuda said:
The equipotential surface passes through a point with field intensity electric 10 kV / m at a distance from a point charge generating a field of r1 = 5 cm.
At what distance from the field generating charge it belongs carry out the second equipotential surface to make the potential difference between these surfaces was equal to 100 V.

BvU said:
No. It's illegible:
I translated task from my main language into english. I don't know what exactly is illegible? Could you explain?

Last edited:
I'm getting the same answers you are getting using your approach; ##6 \text{ cm}## if we're at a lower potential and ##4 \text{ cm}## if we're at a higher potential. But I think your approach is extremely crude but it may work for introductory physics classes. For higher level classes you have to keep in mind that ##\vec{E}## is not constant from equipotential surface to another it actually changes

##E = -\frac{\Delta V}{\Delta d}## only works over very small distances. (minus is important here)

##\vec{E} = 10^4## at the first equipotential surface but not the second.

When I do your problem out properly I get ##6.24 \text{ cm}## at the lower potential and ##\left( 4.16 \text{ cm}\right)## at the higher potential. It all depends if ##\Delta V = + 100 \frac{V}{m}## or ##\Delta V = - 100 \frac{V}{m}##.

polibuda
PhDeezNutz said:
I'm getting the same answers you are getting using your approach; ##6 \text{ cm}## if we're at a lower potential and ##4 \text{ cm}## if we're at a higher potential. But I think your approach is extremely crude but it may work for introductory physics classes. For higher level classes you have to keep in mind that ##\vec{E}## is not constant from equipotential surface to another it actually changes

##E = -\frac{\Delta V}{\Delta d}## only works over very small distances. (minus is important here)

##\vec{E} = 10^4## at the first equipotential surface but not the second.

When I do your problem out properly I get ##6.24 \text{ cm}## at the lower potential and ##\left( 4.16 \text{ cm}\right)## at the higher potential. It all depends if ##\Delta V = + 100 \frac{V}{m}## or ##\Delta V = - 100 \frac{V}{m}##.
Thank you for help, but i have still problem with calculating the second electric intesity. Could you help me?

polibuda said:
Thank you for help, but i have still problem with calculating the second electric intesity. Could you help me?

For this problem you don't have to calculate the second field intensity. You just have to be mindful that there are limitations to ##E = \frac{\Delta V}{\Delta d}## and use another approach. You don't have to find the second field intensity but you do have to be careful of assuming that they are the same at different equipotential surfaces.

Are you familiar with the equation for a positive charge

##E = \frac{Q}{4 \pi \epsilon_0 r^2}##?

and

##V = \frac{Q}{4 \pi \epsilon_0 r}##?

Given that we have ##E = 10^4 \frac{V}{m}## at ##r = 0.05 \text{ m}## we can solve for ##Q## (Do this part on your own and tell me what you get)

Plug your answer for ##Q## into the following and tell me what you get.

##V \left( r = 0.05 \text{ m} \right) = \frac{Q}{4 \pi \epsilon_0 \left( 0.05 \right)} ##

then solve the equations

##V \pm 100 = \frac{Q}{4 \pi \epsilon_0 r_2}##

for ##r_2## that will tell you the distance from the original point charge to the second equipotential surface.

Like I said, depending on what level you're at the answers ##4 \text{cm}## and ##6 \text{cm}## may be right.

PhDeezNutz said:
For this problem you don't have to calculate the second field intensity. You just have to be mindful that there are limitations to ##E = \frac{\Delta V}{\Delta d}## and use another approach. You don't have to find the second field intensity but you do have to be careful of assuming that they are the same at different equipotential surfaces.

Are you familiar with the equation for a positive charge

##E = \frac{Q}{4 \pi \epsilon_0 r^2}##?

and

##V = \frac{Q}{4 \pi \epsilon_0 r}##?

Given that we have ##E = 10^4 \frac{V}{m}## at ##r = 0.05 \text{ m}## we can solve for ##Q## (Do this part on your own and tell me what you get)

Plug your answer for ##Q## into the following and tell me what you get.

##V \left( r = 0.05 \text{ m} \right) = \frac{Q}{4 \pi \epsilon_0 \left( 0.05 \right)} ##

then solve the equations

##V \pm 100 = \frac{Q}{4 \pi \epsilon_0 r_2}##

for ##r_2## that will tell you the distance from the original point charge to the second equipotential surface.

Like I said, depending on what level you're at the answers ##4 \text{cm}## and ##6 \text{cm}## may be right.
Thanks for help. Now I know how to solve this task. You helped me so much :) Thank you again :)

PhDeezNutz
polibuda said:
Thanks for help. Now I know how to solve this task. You helped me so much :) Thank you again :)

Happy to help! Let me know what you get.

PhDeezNutz
That looks good to me.

Thank you very much. Solution must be correct :)

polibuda said:
I translated task from my main language into english. I don't know what exactly is illegible? Could you explain?
It is perfectly legible. The problem is that it is unintelligible, i.e. the English is so broken that it is impossible to decipher its meaning.

Based on your attempt, I think you meant
"there is a (positive?) point charge, and at radius 5cm from it the field is 10kV/m. At some greater (lesser?) distance, the potential is 100V less (more?). What is that distance?"

But I still cannot fit that to the text in post #1.

BvU

## 1. What is electrostatic potential?

Electrostatic potential is a measure of the potential energy of a charged particle in an electric field. It is also known as electric potential or voltage.

## 2. How is electrostatic potential related to electric field?

Electrostatic potential is directly related to electric field. The electric field at a point is equal to the negative gradient of the electrostatic potential at that point.

## 3. What is the formula for calculating electrostatic potential due to a point charge?

The formula for calculating electrostatic potential due to a point charge is V = kQ/r, where V is the potential, k is the Coulomb's constant, Q is the charge of the point charge, and r is the distance from the point charge.

## 4. How does the distance from a point charge affect the electrostatic potential?

The electrostatic potential decreases as the distance from a point charge increases. This is because the electric field strength decreases with distance and the potential is directly related to the electric field.

## 5. Can electrostatic potential be negative?

Yes, electrostatic potential can be negative. This occurs when the potential energy of a charged particle is lower than the reference point, which is typically taken to be infinity.

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