# Calculating Work Required to Move a Point Charge of +8μC from (0,4m) to (3m,0)

• jan2905
In summary, the problem asks how much work is required to move a charge of +8 microC from the point (0,4m) to the point (3m,0) with a fixed point charge of +1mC at the origin. The electric potential is related to the electric field and separation between charges. The equation for electric potential is V = (F*d)/q, where F is the electric field and d is the separation between charges. The electric field is not uniform, so integration must be used to find the potential difference between the two points. The work done on the charge is equal to the potential difference multiplied by the charge. Therefore, the work required to move the charge is (V2-V1)q
jan2905
A point charge, Q=+1mC, is fixed at the origin. How much work is required to move a charge, q=+8(micro)C, from the point (0,4meters) to the point (3meters,0)?

I was unable to derive anything.

I guessed at around 40J. Correct? If so HOW! lol

How are electric potential and work related?

Potential=E*d

You have there an equation relating the electric potential to the electric field and the separation between the change in potential. While it's useful you need to find a way to relate V to W...

I don't know... V=Eq*d ? because F=Eq ?

Ok so it V = E * d then it seems logical to say that

V = (F*d)/q, right? Now the nice thing about electric potential is that it is a scalar.

Last edited:
Since the electic field created by Q is not uniform E(x) must be used

$$E(x)=K\frac{Q}{x^2}$$

where K is

$$\frac{1}{4\pi\varepsilon_0}$$

The force on q is

$$F(x)=qE(x)$$

So, the work done on q is

$$W=-\int\mbox{F(x)dx}$$

This problem doesn't require integration.

and... how did you derive V = (F*d)/q? if F=Eq, then by your derivation V=(Eq)*d/q, which is reduced to V=Ed? how does V=Eqd=Ed?

Typing error... V=E*d => V = F*d / q

Integration is a way to solve the problem. Simply choose a convenient path starting at (0,4) and ending at (3,0). Any path will work because E is a conservative field. Using potential is another way. Finding the potential difference between the two points then multiplying this difference by q will give the desired result. Because E is not uniform use integration.

$$V=-\int_{\infty}^{r}\mbox{E(r)dr}=\frac{Q}{4\pi \varepsilon_0\mbox{r}}$$

$$V_1=\int_{4}^{\infty}\mbox{E(y)dy}=\frac{Q}{4\pi \varepsilon_0\mbox{4}}$$

$$V_2=\int_{3}^{\infty}\mbox{E(x)dx}}=\frac{Q}{4\pi \varepsilon_0\mbox{3}}$$

$$W=(V_2-V_1)q$$

## 1. How is work calculated for moving a point charge?

The work required to move a point charge is calculated using the equation W = qΔV, where W is the work, q is the magnitude of the charge, and ΔV is the change in potential energy.

## 2. What is the unit of work in this calculation?

The unit of work in this calculation is joules (J).

## 3. How do you determine the direction of the work in this scenario?

The direction of the work is determined by the direction of the electric field. If the electric field is in the same direction as the movement of the point charge, the work is considered positive. If the electric field is in the opposite direction of the movement, the work is considered negative.

## 4. What is the value of the electric field in this scenario?

The value of the electric field can be determined using the equation E = kq/r^2, where k is the Coulomb's constant (8.99 x 10^9 Nm^2/C^2), q is the charge of the point charge (in Coulombs), and r is the distance between the point charge and the origin.

## 5. Is the work required to move the point charge the same regardless of the path taken?

Yes, the work required to move a point charge is independent of the path taken, as long as the starting and ending points are the same. This is because the work is only dependent on the change in potential energy, which is the same regardless of the path taken.

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