Question #85532

Four particles carrying charges +2q,-2q,+4q and -4q (with q=1.0 nC) are kept at the vectors of a square of side 6.0 cm.Determine the net electric field due to these charged particles at the centre of square.What is the electrostatic force on a particle carrying positive charge of 1.0 nC placed at the centre of square? Please explain in detail.

Expert's answer

Answer on Question #85532, Physics / Electromagnetism

Question:

Four particles carrying charges +2q,−2q,+4q+2\mathrm{q}, -2\mathrm{q}, +4\mathrm{q} and −4q-4\mathrm{q} (with q=1.0 nCq = 1.0 \, \mathrm{nC} ) are kept at the vectors of a square of side 6.0 cm6.0 \, \mathrm{cm} . Determine the net electric field due to these charged particles at the centre of square. What is the electrostatic force on a particle carrying positive charge of 1.0 nC1.0 \, \mathrm{nC} placed at the centre of square? Please explain in detail.

Solution:


As far as for an electric field E‾=∑iEi‾\overline{E} = \sum_{i}\overline{E_{i}} , then in accordance with the figure above


E=2k2q22a2=429⋅236⋅10−4=28.3(kV/m).RespectivelytheforceE = \sqrt {2} \frac {k 2 q \sqrt {2} ^ {2}}{a ^ {2}} = \frac {4 \sqrt {2} 9 \cdot 2}{3 6 \cdot 1 0 ^ {- 4}} = 2 8. 3 (\mathrm {k V / m}). \quad \text {Respectively} \quad \text {the} \quad \text {force}f=Eq=28.3⋅103⋅10−9=28.3(mcN).f = E q = 2 8. 3 \cdot 1 0 ^ {3} \cdot 1 0 ^ {- 9} = 2 8. 3 (\mathrm {m c N}).

The answer:

The net electric field E=28.3kV/m\mathrm{E} = 28.3\mathrm{kV / m}

the electrostatic force f=28.3mcN\mathrm{f} = 28.3\mathrm{mcN}

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