Question

Dos cargas iguales y opuestas de magnitudes 4x10-6 C están separadas 10 cm.
a) ¿Cuál es la magnitud y dirección del campo eléctrico en un punto justo a la mitad de las cargas?
b) ¿Qué fuerza obraría sobre un electrón colocado allí?

Answer 1

Answer:

Part a)

Electric field due to both charges at its mid point is given as

$E = 2.88 \times 10^7 N/C$

Part b)

Force on electron placed at mid point is given as

$F = 4.61 \times 10^{-12} N$

Explanation:

As we know that there is two equal and opposite charges in this configuration

it is given as

$q = 4 \times 10^{-6} C$

they are separated by d = 10 cm

Part a)

Electric field at mid point will be double of electric field due to any one charge given as

$E = 2\frac{kq}{r^2}$

$E = 2\frac{(9\times 10^9)(4\times 10^{-6})}{0.05^2}$

$E = 2.88 \times 10^7 N/C$

Part b)

Now if an electron is placed at mid point of two charges then net force on this electron is given as

$F = eE$

$F = (1.6 \times 10^{-19})(2.88 \times 10^7)$

$F = 4.61 \times 10^{-12} N$

#Learn

Topic : Electric field and electric force

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