Question

Constant Part E A paralel plate capacitor with plate area A-40 m2 and plate separation d - 3.0 mm is connected to a 35-V battery refer to Figure 130 With the capacitor still connected to the battery a slab of plastic with dielectric strength K.32 is placed between the plates of the capacitor, so that the gap is completely filled with the dielectric refer to figure 2) What are the new value of electric field in the capacitor? Express your answer to two significant figures and include the appropriate units. Value Units Submit Best Answer - Part F Figure 2 of 2 What are the new value of the capacitance Express your answer to two significant figures and include the appropriate units. NA 2 Value Units 35 V K3.2 3.0 mm Sum Rest - Part MacBook Pro

Answer 1

Solution & A = 4.0 m2 d = 3.0mm V = 35 volts initial Electric field Eos >lo Electric field = EO - v dk ax - 35 3.00103 x 3.2 Enew 3645-83 N/c 3.6 X10 3 nlc or volt Volt/me metre ço A K 8.85 % x10-12 x4 x 3.2 1) Capacitance X d 3x10-3 3.8 X108 Farad charge = Q = CV = 3.8x10-8 x 35 1-3 815 6 Coulomb 1-3x10-6 x 35 Energy Tous 24** 2-3 x10-5 Joules