Question

Three charges each equal to +4nC are placed at the three corners of a square of side 2cm. Find the electric field at the fourth corner.

Answer 1

Hint: Calculate the electric field due to each charge and do the vector addition of these fields to find the effective electric field. You can use the superposition theorem to solve this problem. After finding the value of the electric field use vector addition to determine the effective electric field at that point.

Formula used:
Electric field due to a point charge at a given point is given by,
\[E=\dfrac{1}{4\pi {{\varepsilon }_{0}}}\dfrac{Q}{{{r}^{2}}}\] .................. (1)
Where, \[Q\] is the charge of the particle.
              \[r\] is the distance between the point and the charge.
             \[{{\varepsilon }_{0}}\] is the permittivity of the space.

Complete answer:
Let’s look at the following diagram to understand the system,

             

Given that,
\[\begin{align}
  & BC=2cm=2\times {{10}^{-2}}m \\
 & CD=2cm=2\times {{10}^{-2}}m \\
 & AC=2\sqrt{2}cm=2\sqrt{2}\times {{10}^{-2}}m \\
\end{align}\]

So, using equation (1) we get,
\[{{E}_{B}}=\dfrac{1}{4\pi {{\varepsilon }_{0}}}\dfrac{Q}{{{r}^{2}}}=(9\times {{10}^{9}})\dfrac{4\times {{10}^{-9}}}{{{(2\times {{10}^{-2}})}^{2}}}=9\times {{10}^{4}}\] - along BC
\[{{E}_{D}}=\dfrac{1}{4\pi {{\varepsilon }_{0}}}\dfrac{Q}{{{r}^{2}}}=(9\times {{10}^{9}})\dfrac{4\times {{10}^{-9}}}{{{(2\times {{10}^{-2}})}^{2}}}=9\times {{10}^{4}}\] - along DC
\[{{E}_{A}}=\dfrac{1}{4\pi {{\varepsilon }_{0}}}\dfrac{Q}{{{r}^{2}}}=(9\times {{10}^{9}})\dfrac{4\times {{10}^{-9}}}{{{(2\sqrt{2}\times {{10}^{-2}})}^{2}}}=4.5\times {{10}^{4}}\] - along AC

As we can see the following electric fields are equal and perpendicular to each other,
According to the vector addition, the resultant vector due to the above two vectors will be along the diagonal AC and will be aligned with \[{{E}_{A}}\] .
Hence, the value of the resultant vector is,
\[{{E}_{BD}}=9\sqrt{2}\times {{10}^{4}}\] - along AC

Now the vectors \[{{E}_{BD}}\And {{E}_{A}}\] are completely aligned with each other.
So, we can simply add these vectors to find the resultant electric field.
Then, the effective electric field at point C will be \[{{E}_{BD}}+{{E}_{A}}\] .
Hence, the answer of the problem is, \[{{E}_{BD}}+{{E}_{A}}=(9\sqrt{2}+4.5)\times {{10}^{4}} = 17.22 \times 10^4 N{{C}^{-1}}\] .

Note:
You could also solve this problem by breaking the component along AC into x and y components. We can add the individual components along the x and y-axis. Then, we can perform the simple vector addition to find the resultant vector in the following manner.
X component of the electric field is given by,
\[{{E}_{x}}={{E}_{D}}+{{E}_{A}}\cos (45)=12.18\times {{10}^{4}}\]
Y component of the electric field is given by,
\[{{E}_{y}}={{E}_{B}}+{{E}_{A}}\sin (45)=12.18\times {{10}^{4}}\]
Hence, the effective electric field is,
\[E=\sqrt{{{E}_{x}}+{{E}_{y}}}=12.18\sqrt{2}\times {{10}^{4}}=17.22\times {{10}^{4}}N{{C}^{-1}}\]
We can follow either of the two methods to determine the effective electric field at the fourth corner. You should be well aware of the direction of the electric field. It is important because the electric field is a vector quantity.