A rodis uniformly charged with positue electrical charge. Calculate the ratio Ed E
F
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of the magnitude of the elecisic field created by the rod at points D and F. Give your answer rounded to two decimal places.
The correct answer and explanation is:
To calculate the ratio of the electric field magnitudes (ED/EFE_D/E_F) created by a uniformly charged rod at two different points (D and F), we first need to define the situation more clearly. The electric field due to a uniformly charged rod at a point along its axis or perpendicular to it depends on the distribution of the charge and the distance from the rod to the point in question.
Assumptions:
- The rod is uniformly charged with a positive charge QQ distributed along its length.
- The rod has a total length LL, and the points D and F are located at different distances from the rod (we assume D and F lie on the axis or perpendicular to the rod).
Let’s assume the electric field is being calculated at two points:
- Point D is at a distance rDr_D from the center of the rod.
- Point F is at a distance rFr_F from the center of the rod.
Electric Field of a Uniformly Charged Rod:
The electric field due to a uniformly charged rod at a point along its axis (for simplicity) is given by the formula: E=14πϵ02kQL(1r)E = \frac{1}{4 \pi \epsilon_0} \frac{2kQ}{L} \left( \frac{1}{r} \right)
where:
- kk is Coulomb’s constant (k=9×109 N\cdotpm2/C2k = 9 \times 10^9 \, \text{N·m}^2/\text{C}^2),
- QQ is the total charge on the rod,
- LL is the length of the rod,
- rr is the distance from the point to the center of the rod.
Ratio of Electric Fields at Points D and F:
Now, the ratio of the electric fields at points D and F will depend on their respective distances from the rod. The formula for the ratio of the magnitudes of the electric fields is: EDEF=1rD1rF=rFrD\frac{E_D}{E_F} = \frac{ \frac{1}{r_D} }{ \frac{1}{r_F} } = \frac{r_F}{r_D}
So, if the distances rDr_D and rFr_F are known, you can directly calculate this ratio.
Example Calculation:
Let’s assume that point D is located at a distance of 2 meters and point F is located at a distance of 4 meters from the center of the rod. The ratio of the electric field magnitudes would be: EDEF=rFrD=42=2\frac{E_D}{E_F} = \frac{r_F}{r_D} = \frac{4}{2} = 2
Therefore, the ratio of the electric fields at points D and F would be 2.
Explanation:
The electric field strength due to a uniformly charged rod decreases with increasing distance from the rod. The field follows an inverse proportionality with distance, meaning that as the distance increases, the electric field becomes weaker. Hence, when comparing the fields at points D and F, the farther point (F) will experience a weaker electric field. Therefore, the ratio of the electric field magnitudes is simply the ratio of their distances from the rod.