If the capacitor were filled with paper o dielectric con at ant K - 3.5, the cap

Question

If the capacitor were filled with paper o dielectric con at ant K - 3.5, the capacitance would be 20pF 70 pF 245 pF Chooser any two problems from the following three. Two point charges q_1 - 3.0 x 10^2 C and q_1 = 4.0 x 10^-6 C are located on the y-axis at y_1 = -0.4 m and y_2 =-0.4 m y_2 = 0.5 respectively. Find the electric field (magnitude and direction) at the origin and the electric force on a test charge of q = 1.2 x 10^-13 C located at the origin (magnitude and direction Two point charges q_1 = 5.0 x 10^6 C and q_2 = -3.0 x 10^-6 C are located on the x-axis at x_1 = -0.3 m and x2 = 0.2 m. respectively. Find the electric potential at a point P on the y-axis where y = 0.4 m, and the potential energy of a test charge of q =1.8 x 10^-12 C at the point P. A parallel-plate capacitor is connected to a 16.0-V battery. The area on each plate is 24 cm^2 and the spacing between the places is 0.015 cm. The space between the plates is filled with paper of dielectric constant 3.4 . Find the capacitance of the capacitor; the electric field between the plates; the electric energy stored in the capacitor.

Explanation / Answer

E1y = +k*q1/y1^2             E1x = 0

E2y = +k*q2/y2^2             E2x = 0

Enet = E1y + E2y

Enet = ((9*10^9*3*10^-6)/(0.4^2)) + ((9*10^9*4*10^-6)/(0.5^2))

Enet = 312750 N/C

direction along +y axis

electric force F = E*q = 312750*1.2*10^-12 = 3.753*10^-7 N

+++++++++++++

B)

V = v1 + v2

v = kq1/r1 + k*q2/r2

r1 = sqrt(x1^2+y^2) = sqrt(0.3^2+0.4^2) = 0.5 m

r2 = sqrt(0.2^2+0.4^2) = 0.45 m

V = (9*10^9*5*10^-4/0.5) - (9*10^9*3*10^-4/0.45)

V = 3*10^6 V

potential energy U = V*q = 3*10^6*1.8*10^-12 = 5.4*10^-6 J

+++++++++++

c)

(a) C = k*eo*A/d = (3.4*8.85*10^-12*24*10^-4)/(0.015*10^-2) = 4.8144*10^-10 F

(b)

E = V/d = 16/(0.015*10^-2) = 10.7*10^4 N/C

(c)

Q = 0.5*C*V^2 = 0.5*(4.8144*10^-10)*16^2 = 6.162432*10^-8 J

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