The concepts used to resolve the problem are the electrical fields due to potential and potential energy. Initially, determine which key is positive by detect the greater potential plate. Then, usage the expression of an electric field to find the electrical field within the capacitor. Then, use the expression that potential energy to uncover the potential energy at the capacitor"s mid-point.

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Fundamentals

The electric field have the right to be offered as follows:

(E=fracDelta VDelta x)

Here, (mathrmE) is the electrical field, (Delta V) is the change in potential, and also (Delta x) is the distance between both potentials. The electric potential power of a charge at potential (mathrmV) can be determined as follows:

(U=q V)

Here, (cup) is the electric potential, and also (q) is the charge present at the potential (V).

(1) The plate having higher potential will certainly be dubbed a confident plate. The plate through (300 mathrm~V) is at higher potential as contrasted to the plate with (0 mathrm~V), such that the plate through (300 mathrm~V) is dubbed as a positive plate. Thus, the best plate is positive.

Part 1 The right plate is the confident plate.

The worth of that potential determines the positivity of a plate contrasted to the other plate. If the very first plate has actually a 0 V potential and the second plate has actually a -8 potential, then the very first plate is positive.

(2) The electrical field can be provided as follows:

(E=fracDelta VDelta x)

Here, (mathrmE) is the electrical field, (Delta V Delta x Delta V) is the readjust in potential, and (Delta x) is the distance between both potentials. The change in potential in situation of this capacitor is together follows:

(Delta V=V_mathrmH-V_mathrmL)

Here, (V_mathrmH) is the greater potential and (V_mathrmL) is the reduced potential. Instead of (300 mathrm~V) because that (V_mathrmH) and also (0 mathrm~V) for (V_mathrmL) in the above expression.

\$\$ eginaligned Delta V &=300 mathrm~V-0 mathrm~V \ &=300 mathrm~V endaligned \$\$

Substitute (300 mathrm~V) for (Delta V) and also (3.0 mathrm~mm) because that (Delta x) in the expression (E=fracDelta VDelta x)

\$\$ E=frac300 mathrm~V3.0 mathrm~mmleft(frac1 mathrm~mm10^-3 mathrm~m ight) \$\$

(=1.0 imes 10^5 mathrm~V / mathrmm)

Part 2 The electric field toughness inside the capacitor is (1.0 imes 10^5 mathrm~V / mathrmm)

The direction of the electrical field inside the capacitor is from greater potential come the lower potential. Thus, the direction of electric field is native plate through 300 V potential come the plate v 0 V potential.

(3) The electrical potential energy of a fee at potential (mathrmV) can be determined as follows:

(U=q V)

Here, (U) is the electrical potential and (q) is the charge existing at the potential (V). The potential at the midpoint of the capacitor lies between potentials (100 mathrm~V) and (200 mathrm~V). Thus, the potential in ~ the midpoint can be thought about as (150 mathrm~V). Substitute (150 mathrm~V) because that (mathrmV) and (1.6 imes 10^-19 mathrmC) for (mathrmq) in the above expression.

\$\$ eginarrayc U=left(1.6 imes 10^-19 mathrmC ight)(150 mathrm~V) \ =2.4 imes 10^-17 mathrm~J endarray \$\$

Part 3 The potential power is (2.4 imes 10^-17 mathrm~J).

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The capacitor"s midpoint is at fifty percent of the full separation in between the two plates of the capacitor. Here, the capacitor has actually a 3 mm separation. Thus, the midpoint lies at a separation the 1.5 mm.