How Is Work Calculated for a Charge Moving in a Radial Electric Field?

[Winzer]

Homework Statement

Carefully study the following plot of electric field E in N/C versus distance r in m and answer the questions that follow. The electric field is directed radially outward, and the variation of E with r is independent of direction.

Refering to the figure, determine the work done by the electric field on a 7.75 C charge moved from A to B.

Homework Equations

$$W=q \int\vec{E}\dot\vec{dr}$$

The Attempt at a Solution

First I found the charge caused by the electric field( I used E=25 and r=2 since they intersect:
$$q=\frac{E r^2}{\epsilon_{o}}\longrightarrow q= q=\frac{25* 2^2}{\epsilon_{o}}= 1.11E^-8<br /> C$$
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I am finding Work done from a to b:
$$W= q\int\vec{E}\dot\vec{dr}$$ I am pretty sure I am integrating with respect to r.
$$W= q\int\frac{kq_{1}}{r^2}dr$$ a=10 b=4
So what do you think?

 

[Winzer]

anyone?

 

[learningphysics]

anyone?

Yes, looks right to me. Although you wrote $$q=\frac{E r^2}{\epsilon_{o}}$$ you meant $$q=\frac{E r^2}{k}$$ right?

Also here:
$$W= q\int\vec{E}\dot\vec{dr}$$

$$W= q\int\frac{kq_{1}}{r^2}dr$$

q is the 7.75C and q1 is the 1.11E-8C (doesn't matter mathematically... but for your substitution to make sense E = kq1/r^2... hence q1 is the charge creating the field)...

just making sure because you used q for 1.11E-8 before...

Everything looks good.

 

[Winzer]

Yes, looks right to me. Although you wrote $$q=\frac{E r^2}{\epsilon_{o}}$$ you meant $$q=\frac{E r^2}{k}$$ right?

Also here:
$$W= q\int\vec{E}\dot\vec{dr}$$

$$W= q\int\frac{kq_{1}}{r^2}dr$$

q is the 7.75C and q1 is the 1.11E-8C (doesn't matter mathematically... but for your substitution to make sense E = kq1/r^2... hence q1 is the charge creating the field)...

just making sure because you used q for 1.11E-8 before...

Everything looks good.

That is Correct i meant K.
So: $$(9.0e^9* 1.11E^-8* 7.75E^-6 * \int \frac{1}{r^2} dr$$ boudries of a=10 to b=4. = -1.161E-4 J, shouldn't work be positive, we are going against the e-field
Oh and by the way the charge is suppose to be 7.75 uC.

 

[learningphysics]

That is Correct i meant K.
So: $$(9.0e^9* 1.11E^-8* 7.75E^-6 * \int \frac{1}{r^2} dr$$ boudries of a=10 to b=4. = -1.161E-4 J, shouldn't work be positive, we are going against the e-field
Oh and by the way the charge is suppose to be 7.75 uC.

It should be negative since we're calculating the work done by the field, not the change in electric potential energy.

The change in electric potential energy is the negative of the work done by the field... the field does negative work... potential energy increases.

 

[Winzer]

Ok i get it,
Thank you