Chapter 21: Problem 33
A current of \(5.00 \mathrm{~mA}\) is enough to make your muscles twitch. Calculate how many electrons flow through your skin if you are exposed to such a current for \(10.0 \mathrm{~s}\).
Chapter 21: Problem 33
A current of \(5.00 \mathrm{~mA}\) is enough to make your muscles twitch. Calculate how many electrons flow through your skin if you are exposed to such a current for \(10.0 \mathrm{~s}\).
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Get started for freeTwo charged objects experience a mutual repulsive force of \(0.100 \mathrm{~N}\) If the charge of one of the objects is reduced by half and the distance separating the objects is doubled, what is the new force?
Suppose the Earth and the Moon carried positive charges of equal magnitude. How large would the charge need to be to produce an electrostatic repulsion equal to \(1.00 \%\) of the gravitational attraction between the two bodies?
From collisions with cosmic rays and from the solar wind, the Earth has a net electric charge of approximately \(-6.8 \cdot 10^{5} \mathrm{C}\). Find the charge that must be given to a \(1.0-\mathrm{g}\) object for it to be electrostatically levitated close to the Earth's surface.
Consider an electron with mass \(m\) and charge \(-e\) located a distance \(r\) from a fixed proton with mass \(M\) and charge \(+e .\) The electron is released from rest. Which one of the following expressions for the magnitude of the initial acceleration of the electron is correct? a) \(a=\frac{2 k e^{2}}{m M r}\) c) \(a=\frac{1}{2} m e^{2} k^{2}\) e) \(a=\frac{k e^{2}}{m r^{2}}\) b) \(a=\sqrt{\frac{2 e^{2}}{m k r}}\) d) \(a=\frac{2 k e^{2}}{m r}\)
A bead with charge \(q_{1}=1.27 \mu \mathrm{C}\) is fixed in place at the end of a wire that makes an angle of \(\theta=51.3^{\circ}\) with the horizontal. A second bead with mass \(m_{2}=3.77 \mathrm{~g}\) and a charge of \(6.79 \mu \mathrm{C}\) slides without friction on the wire. What is the distance \(d\) at which the force of the Earth's gravity on \(m_{2}\) is balanced by the electrostatic force between the two beads? Neglect the gravitational interaction between the two beads
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