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The electric field in a region is given...

The electric field in a region is given by , `E=(A)/(x^(2))hati+Byhatj+Cz^(2)hatk`
The SI units of A ,B and C are , respectively

A

`(Nm^(3))/(C),Vm^(2),(N)/(m^(2)C)`

B

`Vm^2 ,Vm, (N)/(m ^2 C)`

C

`V//m^(2),V//m,(NC)/(m^(2))`

D

`V//m^2 (Nm^1 )/( c ) (N )/(C )`

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The correct Answer is:
To determine the SI units of A, B, and C in the given electric field expression \( E = \frac{A}{x^2} \hat{i} + B y \hat{j} + C z^2 \hat{k} \), we will follow these steps: ### Step 1: Identify the SI unit of the electric field The electric field \( E \) is defined as force per unit charge. The SI unit of electric field is: \[ \text{SI unit of } E = \frac{\text{Force}}{\text{Charge}} = \frac{\text{Newton}}{\text{Coulomb}} = \text{N/C} \] ### Step 2: Analyze the first term \( \frac{A}{x^2} \) For the term \( \frac{A}{x^2} \) to have the same unit as the electric field, we set: \[ \frac{A}{x^2} = E \] Thus, the SI unit of \( \frac{A}{x^2} \) must also be \( \text{N/C} \). Rearranging gives: \[ A = E \cdot x^2 \] Substituting the unit of \( E \): \[ A = \text{N/C} \cdot \text{m}^2 = \text{N} \cdot \text{m}^2/\text{C} \] ### Step 3: Analyze the second term \( By \) For the term \( By \) to have the same unit as the electric field, we set: \[ By = E \] Thus, the SI unit of \( By \) must also be \( \text{N/C} \). Rearranging gives: \[ B = \frac{E}{y} \] Substituting the unit of \( E \): \[ B = \frac{\text{N/C}}{\text{m}} = \text{N}/(\text{C} \cdot \text{m}) \] ### Step 4: Analyze the third term \( Cz^2 \) For the term \( Cz^2 \) to have the same unit as the electric field, we set: \[ Cz^2 = E \] Thus, the SI unit of \( Cz^2 \) must also be \( \text{N/C} \). Rearranging gives: \[ C = \frac{E}{z^2} \] Substituting the unit of \( E \): \[ C = \frac{\text{N/C}}{\text{m}^2} = \text{N}/(\text{C} \cdot \text{m}^2) \] ### Summary of SI Units - The SI unit of \( A \) is \( \text{N} \cdot \text{m}^2/\text{C} \) - The SI unit of \( B \) is \( \text{N}/(\text{C} \cdot \text{m}) \) - The SI unit of \( C \) is \( \text{N}/(\text{C} \cdot \text{m}^2) \) ### Final Answer The SI units of A, B, and C are: - \( A: \text{N} \cdot \text{m}^2/\text{C} \) - \( B: \text{N}/(\text{C} \cdot \text{m}) \) - \( C: \text{N}/(\text{C} \cdot \text{m}^2) \)

To determine the SI units of A, B, and C in the given electric field expression \( E = \frac{A}{x^2} \hat{i} + B y \hat{j} + C z^2 \hat{k} \), we will follow these steps: ### Step 1: Identify the SI unit of the electric field The electric field \( E \) is defined as force per unit charge. The SI unit of electric field is: \[ \text{SI unit of } E = \frac{\text{Force}}{\text{Charge}} = \frac{\text{Newton}}{\text{Coulomb}} = \text{N/C} \] ...
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