If the electric field between the plates has a magnitude of 2.6×104 N/CN/C , what is the potential difference between the plates?
The correct answer and explanation is:
Correct Answer:
The potential difference VV between two plates is given by the formula: V=E⋅dV = E \cdot d
Where:
- VV is the potential difference in volts (V)
- EE is the electric field strength in newtons per coulomb (N/C)
- dd is the distance between the plates in meters (m)
Since the value of E=2.6×104E = 2.6 \times 10^4 N/C is given, the potential difference depends on the distance dd between the plates. Without knowing dd, the potential difference cannot be numerically calculated.
However, if we suppose for example that the plates are separated by 0.01 m (1 cm), then: V=2.6×104 N/C×0.01 m=260 VV = 2.6 \times 10^4 \, \text{N/C} \times 0.01 \, \text{m} = 260 \, \text{V}
Explanation:
The potential difference between two points in an electric field represents the amount of work needed to move a unit charge between those points. In a uniform electric field, such as the one between two parallel conducting plates, the electric field is constant. This means the voltage, or potential difference, is directly proportional to the distance between the plates.
The equation V=E⋅dV = E \cdot d comes from the definition of electric field as the rate of change of electric potential with distance. This relationship assumes the field is uniform and the movement is directly along the field lines.
In practical applications, this relationship allows engineers and physicists to control the energy of charged particles, such as in particle accelerators or capacitors. The greater the electric field strength or the distance, the higher the potential difference.
If the problem does not give the distance between the plates, the best that can be done is to express the answer in terms of dd: V=(2.6×104)⋅dV = (2.6 \times 10^4) \cdot d
This equation lets you calculate the potential once the distance is known.