10C and 20C are separated by a distance d. If the electric field at the location of a charge 10C is `vec(E )`, the field at the location of 20C is

Options

A. `vec(E )//2`
B. `-vec( E)//2`
C. `-vec(E )`
D. `vec(E )`

Answer

To solve the problem, we need to find the electric field at the location of the 20C charge due to the 10C charge and relate it to the electric field at the location of the 10C charge.

### Step-by-Step Solution:

1. **Identify the Charges and Distance**:
- We have two charges: \( Q_1 = 10C \) and \( Q_2 = 20C \).
- The distance between the charges is \( d \).

2. **Calculate the Electric Field at the Location of 10C due to 20C**:
- The electric field \( \vec{E} \) created by a point charge \( Q \) at a distance \( r \) is given by the formula:
\[ \vec{E} = \frac{k \cdot Q}{r^2} \]
- Here, \( k \) is Coulomb's constant, and \( r \) is the distance from the charge creating the field to the point where the field is being measured.
- For the 10C charge, the electric field due to the 20C charge at that location is:
\[ \vec{E}_{10C} = \frac{k \cdot 20C}{d^2} \]

3. **Direction of the Electric Field**:
- The electric field due to the 20C charge at the location of the 10C charge will be directed away from the 20C charge, since it is positive. Thus, we can denote this field as:
\[ \vec{E}_{10C} = -\frac{20k}{d^2} \hat{i} \]
- (assuming the direction towards the right is positive).

4. **Calculate the Electric Field at the Location of 20C due to 10C**:
- Now, we need to find the electric field at the location of the 20C charge due to the 10C charge:
\[ \vec{E}_{20C} = \frac{k \cdot 10C}{d^2} \]
- The direction of this electric field will be towards the 10C charge (since it is positive), which can be denoted as:
\[ \vec{E}_{20C} = \frac{10k}{d^2} \hat{i} \]

5. **Relate the Electric Fields**:
- We can express the electric field at the location of the 20C charge in terms of the electric field at the location of the 10C charge:
\[ \vec{E}_{20C} = \frac{1}{2} \left(-\vec{E}_{10C}\right) \]
- This means that the electric field at the location of the 20C charge is half the magnitude of the electric field at the location of the 10C charge, but in the opposite direction.

### Final Answer:
The electric field at the location of the 20C charge is:
\[ \vec{E}_{20C} = -\frac{1}{2} \vec{E}_{10C} \]