Answer:
1454.54 kJ/h or 0.404 kW
Explanation:
Provided:
The heat being removed by the refrigerator is 800 kJ/h.
The refrigerator's coefficient of performance (COP) is 2.2.
Since the refrigerator operates for one-fourth of the time,
it effectively removes heat, Q = = 3200 kJ/h.
Next, the power consumed by the refrigerator during its operation is calculated as:
= .
= 1454.54 kJ/h
Thus, the fridge uses 1454.54 kJ/h
or 0.404 kW (given 1 kW = 3600 kJ/h).
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"
Learn more about electric fields:
Answer:
0.018 J
Explanation:
The work required to bring the charge from infinity to the point P is equal to the change in its electric potential energy. This can be expressed as
where
represents the charge's magnitude
and signifies the potential difference between point P and infinity.
After substituting into the formula, we arrive at