solve 20) Four point charges of magnitude q1 = +4.00 pC q-+4.00 pC q-+4.00 C, an

Question

solve 20) Four point charges of magnitude q1 = +4.00 pC q-+4.00 pC q-+4.00 C, and q-4.00 C are located at the corners of a square of side d 0.200 m as shown in the figure above. Determine the resultant force acting on q2- +4.00 HC charge. A) 6.89 N at 225 B) 3.29 N at 45.0 C) 1.49 N at 225° D) 6.89 N at 45.0° E) 8.70 N at 45.0° 21) Three capacitors are connected as shown in the figure above. Their values are C1-2.0 uF C2 = 40pF, and C3 = 40pF What is the equivalent capacitance between points A and B? A) 2.0 F B)1.6F D) 1.0 HF E) 4.0 F

Explanation / Answer

F12 force exerted by q1 on q2 = F12 = k*q1*q2/r12^2

r12 = distance between q1 & q2

F12x = (9*10^9*4*10^-6*4*10^-6)/(0.2^2) = 3.6 N

F12y = 0

F32 force exerted by q3 on q2 = F32 = k*q3*q2/r12^2

r32 = distance between q1 & q2 = d = 0.2

F32y = (9*10^9*4*10^-6*4*10^-6)/(0.2^2) = 3.6 N

F32x = 0

F42 force exerted by q1 on q2 = F42 = k*q4*q2/r12^2

r42 = distance between q1 & q2 = sqrt(2)*d = 1.414*0.2 = 0.2828 m

F42x = (9*10^9*4*10^-6*4*10^-6*cos45)/(0.2828^2) = 1.27 N

F42y = (9*10^9*4*10^-6*4*10^-6*sin45)/(0.2828^2) = 1.27 N

Fx net = F12x + F32x + F42 = 1.27+3.6 = 4.87 N

Fy net = F12y + F32y + F42y = 4.87 N

F2 = sqrt(Fx^2+Fy^2) = 6.898 N

direction = tan^-1(Fy/Fx) = 45

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21)

here C2 & C3 are in parallel

equivalent capitance of 1 & # = c23 = C2 + c3 = 4+4 = 8 uF

c1 & C 23 are in series

equivalent capacitance between a & b = Ceq = C1*C23/(C1+C23)

Cequ = (2*8)/(2+8) = 1.6 uF   <-----------answer

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