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Use English for explanation let k be 9×109

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2 parts
a

399.6 kV\text{399.6 kV}

b

3996 J\text{3996 J}

Step-by-step solution

5 steps

Okay, Engineer, let's re-solve this problem using the specified value for kk and English explanations.

From the diagram and problem description:

  • Side of the square, a=0.5 ma = 0.5 \, \text{m}
  • Charge at P, QP=−22.2 μC=−22.2×10−6 CQ_P = -22.2 \, \mu C = -22.2 \times 10^{-6} \, C
  • Charge at R, QR=+44.4 μC=+44.4×10−6 CQ_R = +44.4 \, \mu C = +44.4 \times 10^{-6} \, C
  • Coulomb's constant, k=9×109 N m2/C2k = 9 \times 10^9 \, \text{N m}^2/\text{C}^2 (as requested)

The distance from S to P (rPSr_{PS}) is the side length of the square, so rPS=0.5 mr_{PS} = 0.5 \, \text{m}.
The distance from S to R (rRSr_{RS}) is also the side length of the square, so rRS=0.5 mr_{RS} = 0.5 \, \text{m}.
The electric potential at S (VSV_S) is the sum of the potentials caused by charge QPQ_P and QRQ_R.

Part a· 3 steps

Calculate Electric Potential at S

  1. Use the formula for electric potential due to a point charge, V=kqrV = \frac{kq}{r}.

    The total potential at S is the sum of potentials from QPQ_P and QRQ_R.

    VS=kQPrPS+kQRrRSV_S = \frac{kQ_P}{r_{PS}} + \frac{kQ_R}{r_{RS}}
  2. Substitute the given values into the formula.

    VS=(9×109 N m2/C2)(−22.2×10−6 C0.5 m++44.4×10−6 C0.5 m)V_S = (9 \times 10^9 \, \text{N m}^2/\text{C}^2) \left( \frac{-22.2 \times 10^{-6} \, C}{0.5 \, \text{m}} + \frac{+44.4 \times 10^{-6} \, C}{0.5 \, \text{m}} \right)
  3. Simplify the calculation.

    VS=(9×109)((−22.2+44.4)×10−60.5) VV_S = (9 \times 10^9) \left( \frac{(-22.2 + 44.4) \times 10^{-6}}{0.5} \right) \, \text{V} VS=(9×109)(22.2×10−60.5) VV_S = (9 \times 10^9) \left( \frac{22.2 \times 10^{-6}}{0.5} \right) \, \text{V} VS=(9×109)(44.4×10−6) VV_S = (9 \times 10^9) (44.4 \times 10^{-6}) \, \text{V} VS=399.6×103 VV_S = 399.6 \times 10^3 \, \text{V} VS=399.6 kVV_S = 399.6 \, \text{kV}

    The electric potential at S is 399.6 kV\boxed{\text{399.6 kV}}.

Part b

3996 J\text{3996 J}

Potential Energy of a 10 mC Charge Placed at S

  1. Use the formula for potential energy, U=q′VU = q'V, where q′q' is the charge placed and VV is the potential at that location.

    The charge placed at S, q′=10 mC=10×10−3 C=1×10−2 Cq' = 10 \, \text{mC} = 10 \times 10^{-3} \, C = 1 \times 10^{-2} \, C.
    The potential at S, VS=399.6×103 VV_S = 399.6 \times 10^3 \, \text{V}.

    US=q′VSU_S = q' V_S US=(1×10−2 C)×(399.6×103 V)U_S = (1 \times 10^{-2} \, C) \times (399.6 \times 10^3 \, \text{V})
  2. Perform the calculation.

    US=3996 JU_S = 3996 \, \text{J}

    The potential energy of the 10 mC charge placed at S is 3996 J\boxed{\text{3996 J}}.

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