A small segment of wire contains 10 nC of charge. The segment is shrunk to one-third of its original length. A proton is very fa
1 answer:
To address this query, we will utilize principles related to electric fields, linear charge density, and electrostatic forces.
The electric field is described as:
In this context:
= Linear charge density
= Permittivity of free space
r = Distance
The expression for linear charge density can be articulated as:
It is defined as
By substituting,
The initial and final electric force can be represented based on charge and electric field through
Upon substituting the electric field value, we obtain:
If the length is reduced to one-third, then
The electric force ratio becomes
Hence, the determined ratio is 3.
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1) For x = 6.6 cm,
2) For x = 6.6 cm,
3) For x = 1.45 cm,
4) For x = 1.45 cm,
5) Surface charge density at b = 4 cm:
6) At x = 3.34 cm, the x-component of the electric field equals zero
7) Surface charge density at a = 2.9 cm:
8) None of these regions
Explanation:
1)
The electric field from an infinite charge sheet is perpendicular to it:
where
is the surface charge density
represents vacuum permittivity
Outside the slab, the electric field behaves like that of an infinite sheet.
Consequently, the electric field at x = 6.6 cm (situated to the right of both the slab and sheet) results from the combination of the fields from both:
where
The field from the sheet points left (negative, inward), and the slab’s field points right (positive, outward).
Thus,
and the negative sign indicates a rightward direction.
2)
Both the sheet’s and slab’s fields are perpendicular to their surfaces, directing along the x-axis, hence there's no y-component for the total field.
<pThus, the y-component totals zero.This happens because both the sheet and slab stretch infinitely along the y-axis. Choosing any x-axis point reveals that the y-component of the field, generated by a surface element dS of either the sheet or slab, , will be equal and opposite to the corresponding component from the opposite side,
. Thus, the combined y-direction field is always zero.
3)
This scenario resembles part 1), but the point here is
x = 1.45 cm
which lies between the sheet and the slab. The fields from both contribute leftward as the slab has a negative charge (resulting in an outward field). Thus, the total field computes to
Replacing with expressions from part 1), we get
where the negative illustrates a leftward direction.
4)
This portion parallels part 2). Since both fields remain perpendicular to the slab and sheet, no component exists along the y-axis, thus the electric field's y-component is zero.
5)
Notably, the slab behaves as a conductor, signifying charge mobility within it.
The net charge on the slab is positive, indicating a surplus of positive charge. With the negatively charged sheet on the left of the slab, positive charges shift towards the left slab edge (at a = 2.9 cm), while negative charges move to the right edge (at b = 4 cm).
The surface charge density per unit area of the slab is
This average denotes the surface charge density on both slab sides at points a and b:
(1)
Additionally, the infinite sheet at x = 0 negatively charged , induces an opposite net charge on the slab's left surface, thus
(2)
Having equations (1) and (2) allows for solving the surface charge densities at a and b, yielding:
6)
We aim to compute the x-component of the electric field at
x = 3.34 cm
This point lies inside the slab, bounded at
a = 2.9 cm
b = 4.0 cm
In a conducting slab, the electric field remains at zero owing to charge equilibrium; thus, the x-component thereof in the slab is zero
7)
From part 5), we determined the surface charge density at x = a = 2.9 cm is
8)
As mentioned in part 6), conductors have zero electric fields internally. Since the slab is conductive, the electric field inside remains zero; therefore, the regions where the electric field is null are
2.9 cm < x < 4 cm
Thus, the suitable answer is
"none of these regions"
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