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Nina [5.8K]
3 years ago
11

A +12 μC charge and -8 μC charge are 4 cm apart. Find the magnitude and direction of the E-field at the point midway between t

he two charges. Use k = 9 x 10^9 N*m^2/C^2.
Physics
1 answer:
Natasha_Volkova [10]3 years ago
8 0

Answer:

Explanation:

Given

Charge of first Particle q_1=+12\ \mu C

Charge of second Particle q_2=-8\ \mu C

distance between them d=4\ cm

k=9\times 10^{9}

magnetic field due to first charge at mid-way between two charged particles is

E_1=\frac{kq_1}{r^2}

r=\frac{d}{2}=\frac{4}{2}=2\ cm

E_1=\frac{9\times 10^9\times 12\times 10^{-6}}{(2\times 10^{-2})^2}

E_1=27\times 10^7\ N/C (away from it)

Electric field due to q_2=-8\ \mu C

E_2=\frac{kq_2}{r^2}

E_2=-\frac{9\times 10^9\times 8\times 10^{-6}}{(2\times 10^{-2})^2}

E_2=-18\times 10^7\ N/C(towards it)

E_{net}=E_1+E_2

E_{net}=9\times 10^7\ N/C(away from first charge)        

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