# Is the Electric Field Calculation Consistent with the Potential Result?

• Cactus
In summary, the conversation discusses the correct way to write and calculate the electric field and potential in part C and D of a question. It is pointed out that the d in the book's scheme and notes are different, and the result for electric field in part C is in agreement with the result for potential in the book. The mistake with d and d' is acknowledged and fixed, and confirmation on the answers is received from a friend.
Cactus
Homework Statement
Hey, I just wanted to post these two questions to check whether the answers I've gotten are correct, as I have no real way of checking if what I've done is correct (This is for parts c and d of the attached question)

For c, would the electric field along the z axis just be the superposition of the field from a negatively charged plate plus the field of a positively charged sphere (as along the z axis the field lines are parallel to the axis)

Likewise, for d, would this be the correct way to go about solving for capacitance given the formula for potential, as I can't see any other way to cancel the Vo and express capacitance purely in geometric terms
Relevant Equations
Capacitance = Charge/Potential Difference
Electric Field of an Infinite Plate
Electric Field of a Sphere
Question

Part C

Part D

at part C How you can write $$E_{-}=\frac{|\pi|}{2\epsilon_0}$$ and what is $$\pi$$

Masano Hutama said:
at part C How you can write $$E_{-}=\frac{|\pi|}{2\epsilon_0}$$ and what is $$\pi$$
ah that's not pi its just a badly written n as in surface charge density

Part c must be correct from you, only thing I have to say is that the d in the book scheme and the d in your notes are different. If I call d' the d of your notes and simply d the d from your book it is

d=d'+R, where R is the radius of the solid sphere,

and having that in mind, the result you get for electric field E in part c seems to be in agreement with the result for potential V that is given by the book(Just integrate your E-field and you ll get the V as presented by the book, we know that ##V=\int Edr##).

Part d also seems correct.

Delta2 said:
Part c must be correct from you, only thing I have to say is that the d in the book scheme and the d in your notes are different. If I call d' the d of your notes and simply d the d from your book it is

d=d'+R, where R is the radius of the solid sphere,

and having that in mind, the result you get for electric field E in part c seems to be in agreement with the result for potential V that is given by the book(Just integrate your E-field and you ll get the V as presented by the book, we know that ##V=\int Edr##).

Part d also seems correct.
Yeah I realized my mistake with the d and d' after posting this and fixed that, but thanks for the reply and confirmation on answers. I've also had a friend finish this question now and got the same answers so fingers crossed they're right

## 1. What is an electric field?

An electric field is a physical field that surrounds electrically charged particles and exerts a force on other charged particles within the field. It is represented by a vector quantity that describes the direction and magnitude of the force.

## 2. How is an electric field created?

An electric field is created by any object that carries an electric charge. This can be due to the presence of excess electrons (negative charge) or a deficiency of electrons (positive charge).

## 3. What is capacitance?

Capacitance is the ability of a system to store an electric charge. It is defined as the ratio of the electric charge stored on a conductor to the potential difference (voltage) between the conductors.

## 4. How is capacitance calculated?

The capacitance of a system can be calculated by dividing the electric charge by the potential difference between the conductors. It is also affected by the distance between the conductors and the type of material used.

## 5. What is the relationship between electric field and capacitance?

The electric field and capacitance are closely related. The electric field strength is directly proportional to the capacitance and inversely proportional to the distance between the conductors. A larger capacitance means a stronger electric field, and a smaller capacitance means a weaker electric field.

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