A charg e ` + 10^9 C` is located at the oergin in free space `&` another charge ` Q` at ` (2,0,0)` . If the X-component of the electric field at (3,1,1) is zero , calculate the value of `Q`, Is the Y-component zero at (3,1,1) ?

Explanation

To solve the problem step by step, we need to calculate the electric field components at the point (3, 1, 1) due to the two charges and determine the value of charge \( Q \) such that the X-component of the electric field is zero. We will also check if the Y-component is zero.

### Step 1: Identify the charges and their positions
- Charge \( q_1 = +10^{-9} \, C \) is located at the origin \( (0, 0, 0) \).
- Charge \( Q \) is located at \( (2, 0, 0) \).

### Step 2: Calculate the position vectors
- The position vector \( \vec{r_1} \) from charge \( q_1 \) to the point \( (3, 1, 1) \):
\[ \vec{r_1} = (3 - 0) \hat{i} + (1 - 0) \hat{j} + (1 - 0) \hat{k} = 3\hat{i} + 1\hat{j} + 1\hat{k} \]

- The position vector \( \vec{r_2} \) from charge \( Q \) to the point \( (3, 1, 1) \):
\[ \vec{r_2} = (3 - 2) \hat{i} + (1 - 0) \hat{j} + (1 - 0) \hat{k} = 1\hat{i} + 1\hat{j} + 1\hat{k} \]

### Step 3: Calculate the magnitudes of the position vectors
- Magnitude of \( \vec{r_1} \):
\[ |\vec{r_1}| = \sqrt{3^2 + 1^2 + 1^2} = \sqrt{9 + 1 + 1} = \sqrt{11} \]

- Magnitude of \( \vec{r_2} \):
\[ |\vec{r_2}| = \sqrt{1^2 + 1^2 + 1^2} = \sqrt{1 + 1 + 1} = \sqrt{3} \]

### Step 4: Calculate the electric fields due to each charge
The electric field \( \vec{E} \) due to a point charge is given by:
\[ \vec{E} = k \frac{q}{r^2} \hat{r} \]
where \( k \) is Coulomb's constant.

- Electric field \( \vec{E_1} \) due to charge \( q_1 \):
\[ \vec{E_1} = k \frac{10^{-9}}{|\vec{r_1}|^2} \hat{r_1} = k \frac{10^{-9}}{11} \left(\frac{3\hat{i} + 1\hat{j} + 1\hat{k}}{\sqrt{11}}\right) \]
\[ \vec{E_1} = k \frac{10^{-9}}{11\sqrt{11}} (3\hat{i} + 1\hat{j} + 1\hat{k}) \]

- Electric field \( \vec{E_2} \) due to charge \( Q \):
\[ \vec{E_2} = k \frac{Q}{|\vec{r_2}|^2} \hat{r_2} = k \frac{Q}{3} \left(\frac{1\hat{i} + 1\hat{j} + 1\hat{k}}{\sqrt{3}}\right) \]
\[ \vec{E_2} = k \frac{Q}{3\sqrt{3}} (1\hat{i} + 1\hat{j} + 1\hat{k}) \]

### Step 5: Set the X-component of the total electric field to zero
The total electric field \( \vec{E} \) at the point \( (3, 1, 1) \) is:
\[ \vec{E} = \vec{E_1} + \vec{E_2} \]

The X-component:
\[ E_x = E_{1x} + E_{2x} = k \frac{10^{-9}}{11\sqrt{11}} \cdot 3 + k \frac{Q}{3\sqrt{3}} \cdot 1 = 0 \]

### Step 6: Solve for \( Q \)
Setting the X-component to zero:
\[ k \frac{10^{-9}}{11\sqrt{11}} \cdot 3 + k \frac{Q}{3\sqrt{3}} = 0 \]
\[ \frac{10^{-9}}{11\sqrt{11}} \cdot 3 + \frac{Q}{3\sqrt{3}} = 0 \]
\[ Q = -\frac{10^{-9} \cdot 3 \cdot 3\sqrt{3}}{11\sqrt{11}} \]

### Step 7: Check the Y-component
The Y-component of the electric field is:
\[ E_y = E_{1y} + E_{2y} = k \frac{10^{-9}}{11\sqrt{11}} \cdot 1 + k \frac{Q}{3\sqrt{3}} \cdot 1 \]
Substituting the value of \( Q \):
\[ E_y = k \frac{10^{-9}}{11\sqrt{11}} + k \frac{-10^{-9} \cdot 3 \cdot 3\sqrt{3}}{11\sqrt{11} \cdot 3\sqrt{3}} = k \frac{10^{-9}}{11\sqrt{11}} - k \frac{10^{-9}}{11\sqrt{11}} = 0 \]

### Conclusion
- The value of \( Q \) is:
\[ Q = -\frac{10^{-9} \cdot 3 \cdot 3\sqrt{3}}{11\sqrt{11}} \]
- The Y-component of the electric field at \( (3, 1, 1) \) is zero.