Problem 102 - A Complete Resource Book in Physics for JEE Main

Question

A point charge \(q\) is placed at origin. Let \(\vec{E}_{A}, \vec{E}_{B}\) and \(\vec{E}_{C}\) be the electric field at three points \(A(1,2,3)\), \(B(1,1,-1)\), and \(C(2,2,2)\) due to charge \(q\). Then (A) \(\vec{E}_{A} \perp \vec{E}_{B}\) (B) \(\vec{E}_{A} \| \vec{E}_{B}\) (C) \(\left|\vec{E}_{B}\right|=4\left|\vec{E}_{C}\right|\) (D) \(\left|\vec{E}_{B}\right|=16\left|\vec{E}_{C}\right|\)

Step-by-Step Solution

Answer

None of the given options (A), (B), (C), or (D) are correct.

Step 1: Define the electric field equation.

The electric field \(\vec{E}\) due to a point charge \(q\) can be calculated using the equation: \[\vec{E} = k \frac{q\vec{r}}{|\vec{r}|^3}\] Where \(k\) is Coulomb's constant, \(\vec{r}\) is the vector from the point charge to the point in space where the electric field is to be measured, and \( |\vec{r}|\) is the distance between the points.

Step 2: Calculate the electric field at point A.

First, let's calculate the electric field at point A: The position vector for point A is: \(\vec{r}_A = \langle 1, 2, 3 \rangle\) Magnitude of position vector for point A: \(|\vec{r}_A| = \sqrt{1^2 + 2^2 + 3^2} = \sqrt{14}\) Now, we can find \(\vec{E}_A\): \[\vec{E}_A = k \frac{q\vec{r}_A}{|\vec{r}_A|^3} = k\frac{q\langle1, 2, 3\rangle}{(14\sqrt{14})}\]

Step 3: Calculate the electric field at point B.

Next, let's calculate the electric field at point B: The position vector for point B is: \(\vec{r}_B = \langle 1, 1, -1 \rangle\) Magnitude of position vector for point B: \(|\vec{r}_B| = \sqrt{1^2 + 1^2 + (-1)^2} = \sqrt{3}\) Now, we can find \(\vec{E}_B\): \[\vec{E}_B = k \frac{q\vec{r}_B}{|\vec{r}_B|^3} = k\frac{q\langle1, 1, -1 \rangle}{(9\sqrt{3})}\]

Step 4: Calculate the electric field at point C.

Now let's calculate the electric field at point C: The position vector for point C is: \(\vec{r}_C = \langle 2, 2, 2 \rangle\) Magnitude of position vector for point C: \(|\vec{r}_C| = \sqrt{2^2 + 2^2 + 2^2} = 2\sqrt{3}\) Now, we can find \(\vec{E}_C\): \[\vec{E}_C = k \frac{q\vec{r}_C}{|\vec{r}_C|^3} = k\frac{q\langle2, 2, 2 \rangle}{(24\sqrt{3})}\]

Step 5: Check each option given the calculated electric fields.

Now that we have electric field expressions for points A, B, and C, we will check the given options: (A): To check if \(\vec{E}_A \perp \vec{E}_B\), the dot product of these two vectors should be equal to zero: \[\vec{E}_A \cdot \vec{E}_B = k^2 q^2 \left(\frac{1}{14\sqrt{14}}\right)\left(\frac{1}{9\sqrt{3}}\right)\left(1\times 1 + 2\times 1 -3\times -1 \right)= \frac{k^2 q^2}{126} (6) \neq 0\] Since the dot product is not equal to zero, option (A) is incorrect. (B): To check if \(\vec{E}_A \| \vec{E}_B\), the vectors must be scalar multiples of each other. We see that the ratios of their components are not equal, so they are not parallel. Hence, option (B) is incorrect. (C): To check if \(\left|\vec{E}_B\right|=4\left|\vec{E}_{C}\right|\), we need to find and compare their magnitudes: \[\frac{|\vec{E}_B|}{|\vec{E}_C|} = \frac{k|q|\frac{|\vec{r}_B|}{|\vec{r}_B|^3}}{k|q|\frac{|\vec{r}_C|}{|\vec{r}_C|^3}}\] \[\frac{|\vec{E}_B|}{|\vec{E}_C|} = \frac{1/9\sqrt{3}}{1/24\sqrt{3}} = \frac{24}{9} = \frac{8}{3} \neq 4\] Since the ratio of their magnitudes is not equal to 4, option (C) is incorrect. (D): To check if \(\left|\vec{E}_B\right|=16\left|\vec{E}_{C}\right|\), let's see if their magnitudes satisfy this condition: \[\frac{|\vec{E}_B|}{|\vec{E}_C|} = \frac{8}{3} \neq 16\] Since the ratio of their magnitudes is not equal to 16, option (D) is also incorrect. None of the given options are correct.