Question

If the charge on an object is doubled then electric field becomes

• A. half
• B. double
• C. unchanged
• D. thrice

Hint:

The electric field measures the force exerted per unit charge. Because it scales linearly with the source charge ($E \propto Q$), any linear change made to the source charge will be mirrored exactly by the electric field强度. If you double the charge, you double the field strength. \nobreak

Answer 1

The Correct Option is B

Concept: The electric field ($\vec{E}$) created by a charged object at any point in space is directly proportional to the magnitude of the source charge ($Q$) creating that field. For a point charge or an object behaving as a point charge, the magnitude of the electric field at a distance $r$ is given by Coulomb's law for electric fields: \[ E = k \cdot \frac{Q}{r^2} \] Where:
• $k$ is the electrostatic constant ($\frac{1}{4\pi\varepsilon_0}$).
• $Q$ is the source charge on the object.
• $r$ is the distance from the center of the charged object to the observation point. From this fundamental relationship, it is evident that keeping the observation distance $r$ constant gives a direct proportionality: \[ E \propto Q \]

Step 1: Set up the initial and final states.
Let the initial charge on the object be $Q_1 = Q$, which produces an initial electric field: \[ E_1 = k \cdot \frac{Q}{r^2} \] According to the problem statement, the new charge on the object is doubled, meaning: \[ Q_2 = 2Q \]

Step 2: Calculate the new electric field ($E_2$).
Substitute the new charge value $Q_2$ into the electric field formula while keeping the distance $r$ exactly the same: \[ E_2 = k \cdot \frac{Q_2}{r^2} = k \cdot \frac{2Q}{r^2} \] Rearranging the terms to factor out the constant coefficient: \[ E_2 = 2 \cdot \left( k \cdot \frac{Q}{r^2} \right) \]

Step 3: Express the final field in terms of the initial field.
Notice that the term inside the parentheses is exactly equal to our initial electric field expression ($E_1$): \[ E_2 = 2 \cdot E_1 \] Therefore, if the charge on the object is doubled, the resulting electric field also becomes double its original value.