I only need the answer to question 6 in this picture. Explain why electric field

Question

I only need the answer to question 6 in this picture.

Explain why electric field lines are always perpendicular to equipotential surfaces. You are going to assemble four equal charges, one at a time, by putting them at. the corners of a square. What is the total work done to assemble these charges if each charge is 5.0 mu C and the square has 25cm sides. Two 1 mu C charges are separated by 2 meters. What is the potential at each charge due to the other charge? What is the potential energy of the two charges? The plates of a capacitor are connected to a. battery. What happens to the charge on the plates if the connecting wires are removed from the battery? What happens to the charge if the plates are now pulled apart? What changes, if any, occur in the energy stored in the capacitor? Two capacitors have an equivalent capacitance of C_p when connected in parallel and C6 when connected in series. What is the capacitance of each capacitor? Find the equivalent capacitance of the capacitor network shown below (Figure 1). Assuming that the potential difference between a and b is 60.0 V and that C_1 = 5.00 mu F, C_2 = 10.00 mu F and C_3 = 2.00 mu F, what is the charge stored on each capacitor?

Explanation / Answer

for the upper branch both c1 and c2 are in series

so

c12 = c1c2/c1+c2 = 3.33*10^-6 F

c12,c12 and c3 are in parallel = c12+c12+c3 = 3.33+3.33+2 = 8.66*10^-6 F

lower branch c2,c2 are in parallel = c2+_c2 = 20*10^-6 F

upper capacitor and lower capacitor are in series

ctot = 8.66*20/28.66 = 6.04*10^-6 F

q tot = c*v = 6.04*10^-6*60 = 362.4*10^-6 C

in series q same

so lower capacitor and upper capacitor has same q = 362.4*10^-6 C

in the lower branch voltage are = 362.4*10^-6/20*10^-6 = 18.12 V

in parallel voltage same

so charge on each c2 in the lower branch = 181.2*10^-6 C

upper branch:q = 181.2*10^-6

voltage = 181.2*10^-6/8.66*10^-6 = 20.92 V

in paralell v is same so v12 = v3 = 20.92 V

q3 = 41.85*10^-6 C

q12 = q12 = 69.7*10^-6 C

in series charge same

so q1 = q2 = 69.7*10^-6 C

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