If a free electron moves through a potential difference of 1 kV, then the energy gained by the electron is given by
1.6 × 10 -16 J
Calculating Energy Gained by a Free Electron in an Electric Field
This question asks us to determine the amount of energy a free electron gains when it moves through a potential difference of 1 kilovolt (kV). Understanding the relationship between electric potential difference, charge, and energy is key to solving this physics problem.
Understanding Potential Difference and Energy Gain
In physics, the potential difference (\(V\)) between two points is defined as the work done per unit charge to move a charge from one point to the other. Conversely, when a charge moves through a potential difference, it gains or loses energy. The energy gained or lost (\(E\)) by a charge (\(q\)) moving through a potential difference (\(V\)) is given by the formula:
\[E = qV\]where:
- \(E\) is the energy gained or lost (in Joules, J)
- \(q\) is the magnitude of the charge (in Coulombs, C)
- \(V\) is the potential difference (in Volts, V)
For an electron, the charge \(q\) is approximately \(1.6 \times 10^{-19}\) Coulombs. A free electron moving through a potential difference will gain kinetic energy equal to the work done by the electric field if only the electric force is doing work.
Applying the Formula to the Problem
We are given the potential difference \(V = 1 \text{ kV}\). We need to convert this to Volts:
\[1 \text{ kV} = 1 \times 1000 \text{ V} = 1000 \text{ V}\]The charge of a free electron is \(q = 1.6 \times 10^{-19}\) C.
Now, we can calculate the energy gained by the electron using the formula \(E = qV\):
\[E = (1.6 \times 10^{-19} \text{ C}) \times (1000 \text{ V})\] \[E = 1.6 \times 10^{-19} \times 10^3 \text{ J}\]When multiplying powers of 10, we add the exponents:
\[E = 1.6 \times 10^{(-19 + 3)} \text{ J}\] \[E = 1.6 \times 10^{-16} \text{ J}\]Comparing with Options
Let's compare our calculated energy value with the given options:
- Option 1: \(1.6 \times 10^{-19}\) J
- Option 2: \(1.6 \times 10^{-16}\) J
- Option 3: \(1 \times 10^{-19}\) J
- Option 4: \(1 \times 10^{-16}\) J
Our calculated value, \(1.6 \times 10^{-16}\) J, matches Option 2.
Alternative Unit: Electron-Volt (eV)
In atomic and particle physics, energy is often expressed in electron-volts (eV). An electron-volt is defined as the energy gained or lost by a single electron moving through an electric potential difference of one volt. So, the energy gained by an electron moving through a 1 V potential difference is 1 eV.
Since \(1 \text{ eV} = 1.602 \times 10^{-19} \text{ J}\) (approximately the charge of an electron in Coulombs), we can also think of the energy gained in eV.
If an electron moves through a potential difference of 1 kV (1000 V), the energy gained is:
\[E = qV\]Using units where \(q\) is measured in elementary charges (like the electron charge) and \(V\) in Volts, the energy in eV is numerically equal to \(V\):
\[E \text{ (in eV)} = (\text{charge in elementary charges}) \times (\text{potential difference in Volts})\]For an electron (charge = 1 elementary charge magnitude) moving through 1000 V:
\[E = 1 \times 1000 \text{ eV} = 1000 \text{ eV}\]To convert 1000 eV to Joules, we multiply by the conversion factor \(1 \text{ eV} \approx 1.6 \times 10^{-19} \text{ J}\):
\[E = 1000 \text{ eV} \times (1.6 \times 10^{-19} \text{ J/eV})\] \[E = 10^3 \times 1.6 \times 10^{-19} \text{ J}\] \[E = 1.6 \times 10^{(-19 + 3)} \text{ J}\] \[E = 1.6 \times 10^{-16} \text{ J}\]This confirms our earlier calculation in Joules.
Revision Table: Key Concepts
| Concept | Definition/Formula | Units |
|---|---|---|
| Potential Difference (\(V\)) | Work done per unit charge (\(W/q\)) | Volts (V) |
| Charge of an Electron (\(q\)) | Fundamental negative charge | Coulombs (C) |
| Energy Gained (\(E\)) | Change in potential energy or kinetic energy | Joules (J) or electron-volts (eV) |
| Formula relating E, q, V | \(E = qV\) | J = C × V |
| Conversion 1 kV to V | 1 kV = 1000 V | - |
| Conversion 1 eV to J | 1 eV \(\approx\) 1.6 \(\times\) 10-19 J | - |
Additional Information: Electric Potential Energy
When a positive charge moves from a point of higher potential to a point of lower potential, it loses potential energy and gains kinetic energy. Conversely, when a negative charge (like an electron) moves from a point of lower potential to a point of higher potential, it loses potential energy (becomes less negative) and gains kinetic energy.
In this problem, the electron gains energy, meaning it moves from a lower potential to a higher potential through a difference of 1 kV.
The energy gained by the electron is electrical potential energy converted into kinetic energy (assuming no energy loss due to other forces). The change in potential energy (\(\Delta U\)) is related to the charge and potential difference:
\[\Delta U = q \Delta V\]Here, \(\Delta V\) is the potential difference the charge moves through. For an electron moving from potential \(V_1\) to \(V_2\), the change in potential energy is \(U_2 - U_1 = q(V_2 - V_1)\). If \(V_2 > V_1\), the potential difference \(V_2 - V_1 > 0\). Since the electron charge \(q\) is negative, \(\Delta U\) will be negative, meaning the potential energy decreases. The energy gained as kinetic energy is \(-\Delta U = -q(V_2 - V_1) = q(V_1 - V_2)\). If \(V\) in the problem refers to \(V_{higher} - V_{lower}\), then the energy gained is \(|q|V\).
The magnitude of the energy gained is \(|q| \times (\text{Potential Difference})\), which is \(E = qV\) using the magnitude of the charge and the magnitude of the potential difference, as calculated in the solution.
Important Questions from Electric Fields and Gauss' Law
A positive charge +q is placed at the centre of a hollow metallic sphere of inner radius a and outer radius b. the electric field at a distance r from the centre is denoted by E. In this regards, which one of the following statement is correct?
- Two point charges $q_1 \left( {\sqrt {10} {\rm{\mu C}}} \right)$ and $q_2(-18\sqrt{2} {\rm{\mu C}})$ are placed on the x-axis at $x = 0$ m and $x = 4$ m respectively. The electric field (in V/m) at a point $(1, 3)$ m is,
$\left[ {{\rm{Take\;}}\frac{1}{{4{\rm{\pi }}{\epsilon_0}}} = 9 \times {{10}^9}{\rm{N}}{{\rm{m}}^2}{{\rm{C}}^{ - 2}}} \right]$ - Let a total charge $2Q$ be distributed in a sphere of radius $R$, with the charge density given by $\rho(r) = Cr^2$, where $r$ is the distance from the centre. Two charges $A$ and $B$, of $-Q$ each, are placed on diametrically opposite points, at equal distance, '$a$' from the centre. If $A$ and $B$ do not experience any force, then:
The expression for torque '\(\vec{\tau}\)' experienced by an electric dipole of dipole moment '\(\vec{P}\)' in an external uniform electric field '\(\vec{E}\)' is given by :
The surface charge density of a thin spherical shell placed in an air medium is 88.54 c/m2 The intensity of the electric field measured 12 mm outside the shell from the centre of the shell is 5.625 × 101 2 N/C. The thin spherical shell has a radius of:
