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Physics Question on potential energy

ABCDA B C D is a rectangle. At corners B,CB, C and DD of the rectangle are placed charges +10×10−10C,−20×10−12C+10 \times 10^{-10} C , \quad-20 \times 10^{-12} C \quad and 10×10−12C10 \times 10^{-12} C, respectively. Calculate the potential at the fourth corner. (The side AB=4cmA B=4 cm and BC=3cm)B C=3 cm )

A

1.65 V

B

0.165 V

C

16.5 V

D

2.65 V

Answer

1.65 V

Explanation

Solution

Now, potential at AA VA=14πε0qBAB+14πε0⋅qCAC+14πε0⋅qDADV_{A}=\frac{1}{4 \pi \varepsilon_{0}} \frac{q_{B}}{A B}+\frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{q_{C}}{A C}+\frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{q_{D}}{A D} =14πε0[10×10−124×10−2−20×10−125×10−2+10×10−123×10−2]= \frac{1}{4 \pi \varepsilon_{0}}\left[\frac{10 \times 10^{-12}}{4 \times 10^{-2}}-\frac{20 \times 10^{-12}}{5 \times 10^{-2}}+\frac{10 \times 10^{-12}}{3 \times 10^{-2}}\right] =9×109×10−10[104−205+103]= 9 \times 10^{9} \times 10^{-10}\left[\frac{10}{4}-\frac{20}{5}+\frac{10}{3}\right] =9×10−1×116= \frac{9 \times 10^{-1} \times 11}{6} =16.5×10−1=1.65V= 16.5 \times 10^{-1}=1.65 V