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A point charge 0.4nC is located at (2, 3, 3). Find the potential differences between (2, 3, 3)m and (-2, 3, 3)m due to the charge.
  • a)
    2.5
  • b)
    2.6
  • c)
    2.7
  • d)
    2.8
Correct answer is option 'C'. Can you explain this answer?
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Verified Answer
A point charge 0.4nC is located at (2, 3, 3). Find the potential diffe...
Vab = (Q/4πεo)(1/rA) + (1/rB), where rA and rB are position vectors rA = 1m and rB = 4m. Thus Vab = 2.7 volts.
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A point charge 0.4nC is located at (2, 3, 3). Find the potential diffe...
Given:
Charge, q = 0.4nC = 0.4 x 10^-9 C
Position of charge, r = (2, 3, 3) m
Position 1, P1 = (2, 3, 3) m
Position 2, P2 = (-2, 3, 3) m
Permittivity of free space, ε0 = 8.854 x 10^-12 F/m

To find: Potential difference between P1 and P2 due to the charge

Formula:
The potential difference between two points due to a point charge can be calculated using the formula:

V = (1/4πε0) x (q/r)

where,
V = Potential difference
ε0 = Permittivity of free space
q = Charge
r = Distance between the point charge and the point where potential difference is to be calculated

Calculation:
Distance between the charge and P1:
r1 = √[(2-2)^2 + (3-3)^2 + (3-3)^2] = 0

Distance between the charge and P2:
r2 = √[(-2-2)^2 + (3-3)^2 + (3-3)^2] = 4

Using the formula, potential difference between charge and P1 can be calculated as:
V1 = (1/4πε0) x (q/r1) = (1/4πε0) x (0.4 x 10^-9/0) = infinity

Using the formula, potential difference between charge and P2 can be calculated as:
V2 = (1/4πε0) x (q/r2) = (1/4πε0) x (0.4 x 10^-9/4) = 2.7 V

Therefore, the potential difference between P1 and P2 due to the charge is 2.7 V.
Hence, the correct option is (C).
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Community Answer
A point charge 0.4nC is located at (2, 3, 3). Find the potential diffe...
The electric field intensity at a point situated 4 meters from a point charge is 200N/C .if the distance is reduced to 2 meters , the field intensity will be
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it depends on the length of the conductor the capacitance of the line is proportional to the length of the transmission line their effect is negligible on the performance of short having a length less than 80 km and low voltage transmission accidents of the transmission line along with the conductances forms the shunted mittens the conductance and the transmission line is because of the leakage over the surface of the conductor considered a line consisting of two conductors and be each of radius are the distance between the conductors being Des shown in the diagram below minus the potential difference between the conductors and via's work QA charge on conductor QB charge on conductor vvab pencil difference between conductor and the Epsilon minus absolute primitivity QA plus QV = 0 so that QA equals QB - equals DBA equals data equals DB equals our substituting these values and voltage equation we get the capacitance between the conductors is cab is referred to as lying to line capacitance if the two conductors are in VR oppositely charge then the potential difference between them is zero then the potential of each conductor is given by one half bath the capacitance between each conductor and point of zero potential and is capacitive CN is called the capacitance to neut or capacitance to ground capacitance cab is the combination of two equal capacity and VN series thus capacitance to neutral is twice the capacitance between the conductors IE CN equals to Cave the absolute primitivity Epsilon is given by Epsilon equals epsilono Epsilon are where epsilano is the permittivity of the free space and Epsilon or is the relative primitivity of the medium prayer capacitance reactants between one conductor and neutral capacitance of the symmetrical three phase line let a balanced system of voltage be applied to a symmetrical three-phase line shown below the phasor diagram of the three phase line with equilateral spacing is shown below take the voltage of conductor to neutral as a reference phaser the potential difference between conductor and we can be written the similarly potential difference between conductors and sea is on adding equations one and two we get also combining equation three and four from equation 6 and 7 the line to neutral capacitance the capacitance of symmetrical three phase line is same as that of the two wire line Related: Capacitance of Transmission Lines?

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A point charge 0.4nC is located at (2, 3, 3). Find the potential differences between (2, 3, 3)m and (-2, 3, 3)m due to the charge.a)2.5b)2.6c)2.7d)2.8Correct answer is option 'C'. Can you explain this answer?
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A point charge 0.4nC is located at (2, 3, 3). Find the potential differences between (2, 3, 3)m and (-2, 3, 3)m due to the charge.a)2.5b)2.6c)2.7d)2.8Correct answer is option 'C'. Can you explain this answer? for Electrical Engineering (EE) 2024 is part of Electrical Engineering (EE) preparation. The Question and answers have been prepared according to the Electrical Engineering (EE) exam syllabus. Information about A point charge 0.4nC is located at (2, 3, 3). Find the potential differences between (2, 3, 3)m and (-2, 3, 3)m due to the charge.a)2.5b)2.6c)2.7d)2.8Correct answer is option 'C'. Can you explain this answer? covers all topics & solutions for Electrical Engineering (EE) 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A point charge 0.4nC is located at (2, 3, 3). Find the potential differences between (2, 3, 3)m and (-2, 3, 3)m due to the charge.a)2.5b)2.6c)2.7d)2.8Correct answer is option 'C'. Can you explain this answer?.
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