Fig.-2 shows three charged particles located at the vertices of an isosceles tri

Question

Fig.-2 shows three charged particles located at the vertices of an isosceles triangle, where d = 2.00 cm and q = 7.00 mu C. Calculate the total electric potential at point A, the midpoint of the base. Calculate the magnitude of the electric filed at point A due to +q (at the top). Calculate the total electric potential energy of the three charges. Calculate the magnitude of the electric force between the two negative charges at the base An ion accelerated through a potential difference of 60.0 V has its potential energy increased by 1.92 times 10^-17 J. Calculate the charge on the ion. Find the speed of a proton that is accelerated from rest through a potential difference of 120 V. What is the magnitude of the electric field, if the cube has sides of 1.20 cm and the net flux is 4 N.m^2/C The net flux of the cube is 20.0 N.m^2/C, what is the enclosed charge? Give at least three properties of an electric charge:

Explanation / Answer

let

q1 = q2 = -q

q3 = +q

3) potential at point A,

V = V1 + V2 + V3

= k*q1/d1 + k*q2/d2 + k*q3/d3

= 9*10^9*(-7*10^-6)/0.01 + 9*10^9*(-7*10^-6)/0.01 + 9*10^9*(7*10^-6)/sqrt(0.04^2 - 0.01^2)

= -1.10*10^7 volts

4) E = k*q3/d3^2

= 9*10^9*7*10^-6/(0.04^2 - 0.01^2)

= 4.30*10^7 N/C

5) U = k*q1*q2/d12 + k*q2*q3/d23 + k*q3*q1/d31

= k*(-q)*(-q)/d + k*(-q)*q/(2*d) + k*q*(-q)/(2*d)

= k*q^2/d - 2*k*q^2/(2*d)

= 0

6) F12 = k*q1*q2/d^2

= 9*10^9*(-7*10^-6)*(-7*10^-6)/0.02^2

= 1.10*10^3 N

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