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Chapter 22: Problem 15

How much work does it take to move a 50 - \(\mu\) C charge against a \(12-V\) potential difference?

Short Answer

Expert verified
The work done to move the charge against the potential difference is 0.0006 Joules.

Step by step solution

01

Understand the Formula

The first step is to understand the formula used to solve this problem. The formula for work done (W) is given by: \( W = Q \times V \) where 'Q' is the electric charge and 'V' is the potential difference.
02

Convert Units

Now we need to convert the charge from micro Coulombs (\(\mu\)C) into Coulombs (C) for use in our formula. We can do this using the conversion factor \(1 \mu C = 1 \times 10^{-6} C\). So, the electrical charge of 50 \(\mu\)C is equal to \(50 \times 10^{-6}\) C.
03

Calculate Work

Now we can calculate the work. Substituting the values into our formula, we get: \( W = (50 \times 10^{-6} C) \times (12 V) = 0.0006 J \)

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Most popular questions from this chapter

Show that the result of Example 22.8 approaches the field of a point charge for \(x \gg a\). (Hint: You'll need to apply the binomial approximation from Appendix A to the expression \(1 / \sqrt{x^{2}+a^{2}}\) )

A charge of \(3.1 \mathrm{C}\) moves from the positive to the negative terminal of a \(9.0-\mathrm{V}\) battery. How much energy does the battery impart to the charge?

A proton, an alpha particle (a bare helium nucleus), and a singly ionized helium atom are accelerated through a \(100-\mathrm{V}\) potential difference. How much energy does each gain?

Two equal but opposite charges form a dipole. Describe the equipotential surface on which \(V=0\)

A sphere of radius \(R\) carries a nonuniform but spherically symmetric volume charge density that results in an electric field in the sphere given by \(\vec{E}=E_{0}(r / R)^{2} \hat{r},\) where \(E_{0}\) is a constant. Find the potential difference from the sphere's surface to its center.
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