Question

Three point charges +Q, -Q and +2Q are placed at the vertices of an isosceles right angled triangle as shown in . The magnitude of electric field intensity produced by these charges at the middle point of hypotenuse is

• A. (1)
• B. (3)

Answer 1

$\displaystyle |\vec{E}|=\frac{2\sqrt{10}\,kQ}{L^2}$; Option (3).

Answer 2

Solution:

We place the triangle so that the base (hypotenuse) is horizontal. Let the vertices be

• $B=(-L/\sqrt2,\,0)$ with charge $-Q$,
• $C=(L/\sqrt2,\,0)$ with charge $+2Q$,
• $A=(0,L/\sqrt2)$ with charge $+Q$ (this vertex is the 90° vertex).

Then the midpoint of the hypotenuse $BC$ is

$$M=\left(\frac{-L/\sqrt2 + L/\sqrt2}{2},\,0\right)=(0,0).$$

Note that each vertex is at a distance

$$r=\frac{L}{\sqrt2}$$

from $M$.

Now, calculate the field at $M$ due to each charge (using $E=\frac{k|q|}{r^2}$):

1. Due to $+Q$ at $A=(0,L/\sqrt2)$:

   Displacement from $A$ to $M$ is $(0-0,\,0-L/\sqrt2)=(0,-L/\sqrt2)$.
   Magnitude:

   $$E_A=\frac{kQ}{(L/\sqrt2)^2}=\frac{kQ}{L^2/2}=\frac{2kQ}{L^2}.$$

   Direction: vertically downward.

2. Due to $-Q$ at $B=(-L/\sqrt2,0)$:

   For a negative charge the field points towards the charge. The vector from $M$ to $B$ is $(-L/\sqrt2,\,0)$ with unit vector $(-1,0)$.
   Magnitude:

   $$E_B=\frac{kQ}{(L/\sqrt2)^2}=\frac{2kQ}{L^2}.$$

   So, $\vec{E}_B=\left(-\frac{2kQ}{L^2},\,0\right)$.

3. Due to $+2Q$ at $C=(L/\sqrt2,0)$:

   For a positive charge the field is directed away from the charge. The vector from $C$ to $M$ is $(0-L/\sqrt2,\,0-0)=(-L/\sqrt2,0)$ with unit vector $(-1,0)$.
   Magnitude:

   $$E_C=\frac{k(2Q)}{(L/\sqrt2)^2}=\frac{2k(2Q)}{L^2}=\frac{4kQ}{L^2}.$$

   So, $\vec{E}_C=\left(-\frac{4kQ}{L^2},\,0\right)$.

Superposition:

Add the x– and y–components:

• $x$–component: $$E_x = E_{Bx} + E_{Cx} = -\frac{2kQ}{L^2} -\frac{4kQ}{L^2} = -\frac{6kQ}{L^2}.$$
• $y$–component: $$E_y = E_{Ay} = -\frac{2kQ}{L^2}.$$

Thus, the net field is

$$\vec{E}=\left(-\frac{6kQ}{L^2},\, -\frac{2kQ}{L^2}\right).$$

Its magnitude is

$$|\vec{E}|=\sqrt{\left(\frac{6kQ}{L^2}\right)^2+\left(\frac{2kQ}{L^2}\right)^2} =\frac{2kQ}{L^2}\sqrt{9+1} =\frac{2\sqrt{10}\,kQ}{L^2}.$$

Since the options given list (1) (3) etc., the correct answer corresponds to option (3).

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Summary:

• Explanation:
  Place the triangle so that the midpoint of the hypotenuse is at the origin. Compute the electric fields from charges at each vertex (noting the directions for positive and negative charges). Sum the vector components to find the net field, whose magnitude comes out as $\frac{2\sqrt{10}\,kQ}{L^2}$.

• Answer:
  $\displaystyle |\vec{E}|=\frac{2\sqrt{10}\,kQ}{L^2}$; Option (3).