## Related questions with answers

Three protons are initially very far apart. Calculate the work that must be done in order to bring these protons to the vertices of an equilateral triangle of side

$5.0 \times 10 ^ { - 16 } \mathrm { m }$

.

Solution

VerifiedWhen we bring the first proton, the potential on it will be zero because there are no charges on the vertices of the triangle, so the work required to bring the first proton is zero. Now, after we put the first proton on one of the vertices of the triangle, it will produce a potential equal to $(k e/a)$ at each of the other two vertices, so the second proton will experience a potential of $(k e/a)$. Therefore, the work required to bring the second proton can be written as follows

$W_{ 2}=q\Delta V=e\times (\frac{ke}{a}-0)=\frac{ke^{2}}{a}$

When we bring the third proton, there will be two potentials due to the two other protons, so the work required to bring the third proton is

$W_{ 3}=q\Delta V=e\times (2\frac{ke}{a}-0)=\frac{2ke^{2}}{a}$

So the total work required to bring the three protons is

$W_{\text{net}}=W_{1}+W_{2}+W_{ 3}=\frac{3ke^{2}}{a}$

$W_{\text{net}}=\frac{3\times (8.99\times 10^{9} \mathrm{~ N\cdot m^{2}/C^{2}}) \times (1.6 \times 10^{-19} \mathrm{~ C})^{2}}{5\times 10^{-16} \mathrm{~ m}}$

$W_{\text{net}}=1.38 \times 10^{-12} \mathrm{~ J}$

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