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hw05 - Physics 21 Spring 2011 Solution to HW-5 24-2 The...

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Physics 21 Spring, 2011 Solution to HW-5 24-2 The electric potential difference between the ground and a cloud in a particular thunderstorm is 1 . 7 × 10 9 V. What is the magnitude of the change in the electric potential en- ergy (in eV) of an electron that moves between the ground and the cloud? Electric potential energy difference for a particle can be cal- culated using the following relation: Δ U = q Δ V where q is the charge of the particle and Δ V is the elec- tric potential difference. Note the difference between elec- tric potential , and electric potential energy . Briefly, electric potential is unit energy per unit charge, so it can be defined for any point in an electric field independent of any actual charges. For an electron, q = - e , so: Δ U = q Δ V = 1 . 7 × 10 9 eV 24-4 Two large, parallel, conducting plates are 13cm apart and have charges of equal magnitude and opposite sign on their facing surfaces. An electrostatic force of 3 . 2 × 10 15 N acts on an electron placed anywhere between the two plates. (Ne- glect fringing.) (a) Find the electric field (in N C ) at the posi- tion of the electron. (b) What is the potential difference in volts between the plates? Equation 22-1 tells us that the electric field due to some electrostatic force is E = F q so plugging in (a) E = F e = 3 . 2 × 10 15N 1 . 602 x 10 19 C = 2 . 00 × 10 4 N C Equation 24-18 tells us that the potential difference between two points is found with the equation Δ V = - integraltext E · d s since the electric field is constant between the plates, we can write Δ V = E Δ s or the potential difference equals the electric field times the seperation between the plates. Plugging in gives us (b) Δ V = parenleftbigg 2 . 00 × 10 4 N C parenrightbigg * 0 . 13m = 2600V 24-9 An infinite, nonconducting sheet has a surface charge density σ = +8 . 14pC / m 2 .
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