A point charge q1- is held stationary at the origin- A second charge q2 is placed at a point a- and the electric potential energy of the pair of charges is -6-4 - 10-8 J- When the second charge is moved to point b- the electric force on the charge does 4-2 - 10-8 J of work- What is the electric potential energy of the pair of charges when the second charge is at point b-

A particle of charge q₁ = 3 μC is located on x-axis at the point x₁ = 6 cm. A second point charge q₂ = 2 μC is placed on the x-axis at x₂ = -4 cm. The absolute electric potential at the origin is

Q. A point charge

q_{1} = - 5.8 μ C

is held stationary at the origin. A second charge

q_{2} = + 4.3 μ C

moves from the point

(0.26 m, 0, 0)

(0.38 m, 0, 0)

. Find the workdone by the electric force on

q_{2}

Q. Two charges of magnitude -4C and 4C are placed at points (1, 2, 3) and (3, 4, 2) respectively. What is the dipole moment?

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The correct option is C −10.6×10−8JPE1=PE2+work done PE2=PE1−work done =−6.4×10−8−4.2×10−8=−10.6×10−8J

The correct option is C $- 10.6 \times 10^{- 8} J$
$P E_{1} = P E_{2} +$ work done
$P E_{2} = P E_{1} -$ work done $= - 6.4 \times 10^{- 8} - 4.2 \times 10^{- 8} = - 10.6 \times 10^{- 8} J$