Two parallel plates having charges of equal magnitude but opposite sign are sepa

Question

Two parallel plates having charges of equal magnitude but opposite sign are separated by 12.0 cm. Each plate has a surface charge density of 36.0 nC/m2. A proton is released from rest at the positive plate. Determine a) the potential difference between the plates, b) the energy of the proton when it reaches the negative plate, c) the speed of the proton just before it strikes the negative plate, d) the acceleration of the proton, and e) the force on the proton. f) From the force, find the magnitude of the electric field and show that it is equal to that found form the charge densities on the plates

Explanation / Answer

distance between plates is d = 12.0 cm = 12.0 * 10^-2 m surface charge density is S = 36.0 nC/m^2 = 36.0 * 10^-9 C/m^2 (S ------ sigma) the electric field between the plates is E = (S/e_o) where e_o = 8.85 * 10^-12 C^2/Nm^2 the potential difference between the plates is V = E * d the energy of the proton is U = V * q where q is charge on proton and has value +1.6 * 10^-19 C let the speed of proton be v we know that K = (1/2)mv^2 here,K = U or (1/2)mv^2 = V * q or v = (2V * q/m)^1/2 where m = 1.67 * 10^-27 kg let the acceleation of proton be a we know that v^2 - u^2 = 2ad or a = (v^2 - u^2/2d) where u = 0 the force on proton is F = m * a let the electric field be E we know that F = E * q or E = (F/q)

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