Question

Point charges q1=+2.00μC and q2=−2.00μC are placed at adjacent corners of a square for which the length of each side is 2.50 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. Part A What is the electric potential at point a due to q1 and q2? Express your answer with the appropriate units. Va V a = nothing nothing Request Answer Part B What is the electric potential at point b? Express your answer with the appropriate units. Vb V b = nothing nothing Request Answer Part C A point charge q3 = -2.00 μC moves from point a to point b. How much work is done on q3 by the electric forces exerted by q1 and q2? Express your answer with the appropriate units. Wab W a b = nothing nothing Request Answer

Answer 1

Answer:

a)  V = 0   V, b) V = -2.11 10⁵ V c) W = 42.3  J

Explanation:

It is electrical potential due to point charges is given by

      V = k ∑ qi / ri

Let's start by reducing the quantities to SI units

     q1 = +2.00 μC = 2.00 10⁻⁶ C

     q2 = - 2.00 μC = -2.00 10⁻⁶ C

     q3 = - 2.00  μC = - 2.00 10⁻⁶ C

     L = 2.50 cm = 0.0250 m

Part A

Let's apply the formula to our case

The distance of each load to point a, we can find it using Pythagoras' theorem

      a² = (L /2)²+ (L /2)² = L² 2/4

      a = L /2 √2

      a = 0.025/2 √2

      a = 1.77 10⁻² m

     V = k (q1 / r1 + q2 / r2)

     V = k (2 10⁻⁶ /a  -2 10⁻⁶ /a)

     V = 0   V

Part B

Let's look for the distance to point b

    $r_{b}$² = L² + L² = L² 2

      $r_{b}$ = L √ 2

     $r_{b}$ = 2.5 10⁻² √2

      $r_{b}$ = 3.54 10⁻² m

Let's calculate the potential

     V = k (q1 /  $r_{b}$ + q2 / L)

     V = 8.99 10⁹ (2 10⁻⁶/3.54 10⁻² - 2 10⁻⁶ / 2.50 10⁻²)

     V= 8.99 10⁹ (0.5649 10⁻⁴ - 0.80 10⁻⁴)

     V = -2.11 10⁵ V

Part C

Electrical work is equal to the change in potential energy, so we will calculate the potential energy at each point

      W = ΔU

For point a

     $U_{a}$ = k (q₁q₃ / a + q₂q₃ / a)

    $U_{a}$  = k (2 3 / a - 2 3 / a)

     $U_{a}$  = 0

For point b

    $U_{b}$  = k (q₁q₃ / rb + q₂q₃ / L)

    $U_{b}$  = 8.99 10⁹ (-2 2 10⁻¹² / 3.54 10⁻² + ​​2 2 10⁻¹² / 2.50 10⁻²)

    $U_{b}$  = 8.99 10⁹ (-1.13 + 1.6) 10⁻¹⁰

    $U_{b}$  = 4.23 101 J

   

    W = $U_{b}$  -$U_{a}$  

    W = 42.3  J