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Question

Four equal charges of

5.0μC5.0 \mu \mathrm { C }

are placed at the vertices of a square of side 10 cm. a. Calculate the value of the electric potential at the centre of the square. b. Determine the electric field at the centre of the square. c. How do you reconcile your answers to a and b with the fact that the electric field is the derivative of the potential?

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.a)\textbf{.a)} From the graph below, we can see that the distance between each charge and the point at the center of the square is half the length of the diagonal of the square. We can use the Pythagorean theorem to find the length of the diagonal, as follows

d=(10 cm)2+(10 cm)2=200=102 cmd=\sqrt{(10\mathrm{~ cm})^{2} + (10 \mathrm{~ cm})^{2}}= \sqrt{200}=10\sqrt{2} \mathrm{~ cm}

Therefore, the distance between each charge and the center of the square is [d/2=52[d/2=5\sqrt{2} cm]. Now, since all the charges have the same magnitude and the same separation of the center, the net electric potential at the center is the value of the potential of one charge multiplied by four, which can be written as follows

Vnet=4×kqr=4×(8.99×109 Nm2/C2)×5×106 C52×102 mV_{\text{net}}=4 \times \frac{kq}{r}=4\times \frac{(8.99 \times 10^{9} \mathrm{~ N\cdot m^{2}/C^{2}}) \times 5\times 10^{-6} \mathrm{~ C} }{5\sqrt{2} \times 10^{-2} \mathrm{~ m}}

Vnet=2.54×106 VV_{\text{net}}=2.54 \times 10^{6} \mathrm{~ V}

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