Chapter 4: Q17E (page 778)

Point charges q₁ = +2.00 µC and q₂ = -2.00 µC are placed at adjacent corners of a square for which the length of each side is 3.00 cm. Point a is at the center of the square, and point b is at the empty corner closest to q₂. Take the electric potential to be zero at a distance far from both charges.
(a) What is the electric potential at point a due to q₁ and q₂?
(b) What is the electric potential at point b?
(c) A point charge q₃ = -5.00 µC moves from point a to point b. How much work is done on q₃ by the electric forces exerted by q₁ and q₂? Is this work positive or negative?

Short Answer

Expert verified

The electric potential at a point due to q₁ and q₂ is Zero.
The electric potential at b is -175 KV.
the work done in moving a charge from a to be is -0.875 Joules

Step by step solution

Given Data and concept

q₁ = 2.0010^-6 C

q₂ = - 2.0010^-6 C

q₃ = -5.0010^-6 C

The electric potential at point r due to point charge q can be written as:

$V = \frac{K q}{r}$

Here V is the potential.

Electric Potential due to q1 and q2

(a)

Electric Potential due both charges at a point is given by,

$V_{a} = \frac{k q_{1}}{d} + \frac{k q_{2}}{d} = \frac{k}{d} (2.00 μ C - 2.00 μ c) = 0$

Therefore, the electric potential at a point due to q₁ and q₂ is Zero.

Electric Potential at b

(b)

The electric potential at b is given by,

$V_{b} = \frac{{kq}_{1}}{d_{1}} + \frac{{kq}_{2}}{d_{2}} = k (\frac{q_{1}}{d_{1}} + \frac{q_{2}}{d_{2}}) = 9 \times 10^{9} (\frac{2.0 \times 10^{- 6}}{0.0424} - \frac{2.0 \times 10^{- 6}}{0.03}) = - 1.75 KV$

Therefore, the electric potential at b is -175 KV.

Work done on q3

(c)

The work done on charge q₃ from a to b is given by,

$W = q_{3} (V_{a} - V_{b}) = (- 5 \times 10^{- 6}) (175 \times 10^{3} - 0) = - 0.875 J$

Therefore, the work done moving a charge from a to be is -0.875 Joules.

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Short Answer

Step by step solution

Given Data and concept

Electric Potential due to q1 and q2

Electric Potential at b

Work done on q3

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