Question

3) If the Capacitance of an object is C so if it is charged by moving charge q from the negative to positive terminal the voltage is Remember voltage is ElectricalPotential(Volts) = Energy(Joules) charge(Coulombs) a. Calculate the energy required if a capacitor of capacitance C is charged up to a total charge Q, remembering that the voltage changes as we are adding charge. b. Consider a parallel plate capacitor of large area A and small plate separation t. Assume the electric field within the plates is constant and normal to the plates. For an initially uncharged capacitor, charge is moved from one plate to the other. i. Give and expression for the surface charge density o on the positive plate. ii. Give an expression for the electric field between the plates. iii. Give an expression for the electric potential difference (voltage) across the plates. From this determine the capacitance. iv. Give an expression for the volume between the plates. V. Express the energy density (energy/volume) as a function of the field strength E UE =

Answer 1

S (a) We have Charge capacitor to calculate the of capacitance energy required to to a total charge 200cc Let, during the process of charging the charge on the positive plate is q then the voltage is Veqa= u Now, let's say that we add an extra infinitesimal Charge dq after a. then energy of the capacitor O SE= I dq ( since SE - Venda since the capacitor is charged from oto & the total energy is Eelse becomes - co * 192.dets 90+1). c [ Q² 07] E 4. Tere Note: We denote electric field by E so we'll denote energy by E to avoid any confusion hence (e= a ² 2 с

(6) Now assuming a a and o assuming a parallel plate capacitor of arge plate area t and plate seperation a assuming that electric field is constant and normal to the blates, and charge Q is moved from one plate to another. The surface charge density is defined as Total charge surface area of the blate - (1) (11) Take the following diagram: 9 Ett+++++ - - - since the so the plate area is reses large, we can consides plates to be infinite for small t. the field due to the positive plate is Ex - J (as shown in the fig) aue to negative flate Et 9 &

positive charge is always away and the the charge. th determine expression concept: The direction of electric field due to charge is always away from the charge. that due to negative charge is always towards charge. the sign of charge is used only to le direction, it doesn't appear in the So, not field between the plates, according to the principle of supperposition is E = Et & E or E = 2 + c = E direction, it a of E. t (ii) since the field is constant between the plates, - the electric potential is Ve Ext & It and capacitance, cale =et I since, C Tea or Q = T A 6,0 A 5 €. A

(iv) Volume between the plates és V= area of the plate X distance between them = Axt « Energy density up - or UE = 2/C At or or UE= UE= Q² ZAct since electric field & so Q = to AE do UE SACT G²A²E2 2 Act ar - E²4 act