What charge would be required to electrify a sphere of radius 25 cm, so as to get a surface charge density of `(3)/(pi)Cm^(-2)`?

Options

A. `3/4 C`
B. `2/4 C`
C. `3/2 C`
D. `3 C`

Answer

To find the charge required to electrify a sphere of radius 25 cm to achieve a surface charge density of \(\frac{3}{\pi} \, \text{C/m}^2\), we can follow these steps:

### Step 1: Understand the formula for surface charge density
The surface charge density (\(\sigma\)) is defined as the charge (\(q\)) per unit area (\(A\)):
\[ \sigma = \frac{q}{A} \]

### Step 2: Calculate the area of the sphere
The area (\(A\)) of a sphere is given by the formula:
\[ A = 4\pi r^2 \]
where \(r\) is the radius of the sphere. Given that the radius is 25 cm, we first convert this to meters:
\[ r = 25 \, \text{cm} = 0.25 \, \text{m} \]
Now, we can calculate the area:
\[ A = 4\pi (0.25)^2 = 4\pi (0.0625) = \pi \times 0.25 = \pi \times 0.25 \, \text{m}^2 \]

### Step 3: Substitute the values into the surface charge density formula
We know the surface charge density \(\sigma = \frac{3}{\pi} \, \text{C/m}^2\). We can rearrange the surface charge density formula to find the charge \(q\):
\[ q = \sigma \times A \]
Substituting the values we have:
\[ q = \left(\frac{3}{\pi}\right) \times \left(4\pi (0.25)^2\right) \]
This simplifies to:
\[ q = \left(\frac{3}{\pi}\right) \times \left(4\pi \times 0.0625\right) \]

### Step 4: Simplify the expression
The \(\pi\) in the numerator and denominator cancels out:
\[ q = 3 \times (4 \times 0.0625) = 3 \times 0.25 = 0.75 \, \text{C} \]

### Final Answer
Thus, the charge required to electrify the sphere is:
\[ q = 0.75 \, \text{C} \]

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