Exam 17: Electric Potential

How much work must we do on an electron to move it from point $A$ , which is at a potential of $+ 50 \mathrm {~V}$ , to point $\mathrm { B }$ , which is at a potential of $- 50 \mathrm {~V}$ , along the semicircular path shown in the figure? Assume the system is isolated from outside forces. $\left( e = 1.60 \times 10 ^ { - 19 } \mathrm { C } \right)$

A parallel-plate capacitor with plate separation of $4.0 \mathrm {~cm}$ has a plate area of $6.0 \times 10 ^ { - 2 } \mathrm {~m} ^ { 2 }$ .What is the capacitance of this capacitor if a dielectric material with a dielectric constant of $2.4$ is placed between the plates, completely filling the space? $\left( \varepsilon _ { 0 } = 8.85 \times 10 ^ { - 12 } \mathrm { C } 2 / \mathrm { N } \cdot \mathrm { m } ^ { 2 } \right.$ )

If the electric field between the plates of a given air-filled capacitor is weakened by removing charge from the plates, the capacitance of that capacitor

A proton that is initially at rest is accelerated through an electric potential difference of magnitude $500 \mathrm {~V}$ . How much kinetic energy does it gain? $\left( e = 1.60 \times 10 ^ { - 19 } \mathrm { C } \right)$

A +3.0- $\mu \mathrm { C }$ point charge is initially extremely far from a positive point charge $Q$ . You find that it takes $41 \mathrm {~J}$ of work to bring the $+ 3.0 - \mu \mathrm { C }$ charge to the point $x = 3.0 \mathrm {~mm} , y = 0.0 \mathrm {~mm}$ and the point charge $Q$ to the point $x = - 3.0 \mathrm {~mm} , y = 0.00 \mathrm {~mm}$ . What is $Q$ ? $( k = \ldots )$

If an electron is accelerated from rest through a potential difference of $1500 \mathrm {~V}$ , what speed does it reach? $\left( e = 1.60 \times 10 ^ { - 19 } \mathrm { C } , m _ { \text {electron } } = 9.11 \times 10 ^ { - 31 } \mathrm {~kg} \right)$

An ideal, isolated, air-filled parallel-plate capacitor is not connected to a battery but has equal and opposite charges of $3.9 \mathrm { nC }$ on its plates. The separation between the plates initially is $1.2 \mathrm {~mm}$ , and for this separation the capacitance is $3.1 \times 10 ^ { - 11 } \mathrm {~F}$ . How much work must be done to pull the plates apart until their separation becomes $7.7 \mathrm {~mm}$ ? $\left( \varepsilon _ { 0 } = \right.$ $\left. 8.85 \times 10 ^ { - 12 } \mathrm { C } ^ { 2 } / \mathrm { N } \cdot \mathrm { m } ^ { 2 } \right)$