Two point charges +1 nC and -3 nC are placed in two adjacent vertices of a squar

Question

Two point charges +1 nC and -3 nC are placed in two adjacent vertices of a square with 3 m long sides. Kind electric field (magnitude and direction) and electric potential (magnitude and sign) in the empty vertices. Find total potential energy of the system. Find DC currents (magnitudes and directions) in all resistors of the circuit shown. Find total electric power developing in this circuit. Neglect internal resistances of the batteries. Find voltage between the points A and B. A metal equilateral triangle of resistance 10 ohm and size 20 cm is pushed with a force of 0.2 N along one of its sides into a homogenous magnetic field with constant speed of 15 m/s. Find strength of magnetic field. An optical system is made of a converging lens with focal distance 16 mm and a concave mirror of radius 41 mm. Find distance between the images produced by this system if the object placed at a distance of 6 mm from the lens. The distance between the lens and minor is 63 mm. Draw the i

Explanation / Answer

1) electric field at third vertex = sigma kq/r^2

= 9e9*1e-9/3^2 (-j) + 9e9*(3e-9)/(2*3^2) (i+j)

= - j + 1.5(i+j)

= 1.5 i + 0.5 j   

= sqrt(1.5^2+0.5^2)

= 1.581 N

direction = arctan(0.5/1.5)

= 18.4 degree

electric field at fourth vertex = sigma kq/r^2

= 9e9*1e-9/(2*3^2) (i-j) + 9e9*(3e-9)/(3^2) (j)

= 3 j + 0.5(i-j)

=0.5 i + 2.5j

= sqrt(2.5^2+0.5^2)

= 2.55 N

direction = arctan(2.5/0.5)

= 78.7 degree

potential at third virtex = sigma kq/r

=   9e9*1e-9/3 + 9e9*(-3e-9)/(3*sqrt 2)

= -3.36 V

potential at fourth virtex = sigma kq/r

=   9e9*1e-9/(3*sqrt 2) + 9e9*(-3e-9)/3

= -6.88 V

potential energy = kq1q2/r

= 9e9*1e-9*(-3e-9)/3

= -9*10^-9 J

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