An electric field converges at the origin whose magnitude is given by the expression `E=100r N//C`, where r is the distance measured from the origin.

Options

A. Total charge contained in any spherical volume with its centre at origin negative.
B. Total charge contained at any spherical volume, irrespective of the location of its centre, is negative
C. Total charge contained in a spherical volume of radius 3 cm with its centre at origin equals `3xx10^(-13) C`
D. Total charge contained in a spherical volume of radius 3 cm with its centre at origin has magnitude `3xx10^(-9) C`.

Explanation

To solve the problem, we need to analyze the electric field given by the expression \( E = 100r \, \text{N/C} \), where \( r \) is the distance from the origin. We will derive the total charge contained within a spherical volume centered at the origin.

### Step 1: Understand the Electric Field Expression
The electric field \( E \) is given as a function of distance \( r \) from the origin. The expression indicates that the electric field strength increases linearly with distance from the origin.

### Step 2: Relate Electric Field to Charge
We know from Gauss's law that the electric field \( E \) due to a point charge \( Q \) at a distance \( r \) is given by:
\[ E = \frac{kQ}{r^2} \]
where \( k \) is Coulomb's constant, approximately \( 9 \times 10^9 \, \text{N m}^2/\text{C}^2 \).

### Step 3: Equate the Two Expressions
We can set the two expressions for the electric field equal to each other:
\[ 100r = \frac{kQ}{r^2} \]

### Step 4: Rearrange to Solve for Charge \( Q \)
Multiplying both sides by \( r^2 \) gives:
\[ 100r^3 = kQ \]
Now, solving for \( Q \):
\[ Q = \frac{100r^3}{k} \]

### Step 5: Substitute the Value of \( k \)
Substituting \( k = 9 \times 10^9 \):
\[ Q = \frac{100r^3}{9 \times 10^9} \]

### Step 6: Calculate Charge for a Specific Radius
Let's calculate the charge \( Q \) for a radius \( r = 3 \, \text{cm} = 0.03 \, \text{m} \):
\[ Q = \frac{100(0.03)^3}{9 \times 10^9} \]
Calculating \( (0.03)^3 \):
\[ (0.03)^3 = 0.000027 \, \text{m}^3 \]
Now substituting back:
\[ Q = \frac{100 \times 0.000027}{9 \times 10^9} = \frac{0.0027}{9 \times 10^9} \]
\[ Q = 3 \times 10^{-13} \, \text{C} \]

### Conclusion
The total charge contained within a spherical volume of radius \( 3 \, \text{cm} \) centered at the origin is \( 3 \times 10^{-13} \, \text{C} \).