Question

two points charges q1=+0.2 C and q2=+0.4C are placed 0.1m apart.Caculate the electric field at (a)the midpoint between the charges and (b)a point on the line joining q1 and q2 such that it is 0.05m away from q2 and 0.15m away from q1

Answer 1

Answer:

Part a)

Electric field at mid point of two charges is

$E = 7.2 \times 10^[11} N/C$

Part b)

Electric field at other position due to two charges

$E = 1.52 \times 10^{12} N/C$

Explanation:

Part a)

Here we know that the electric field at mid point of two charges is given as

$E = \frac{kq_2}{r_2^2} - \frac{kq_1}{r_1^2}$

since both charges are of same nature so electric field due to both charges will be in opposite directions

So we have

$E = \frac{(9\times 10^9)(0.4)}{0.05^2} - \frac{(9\times 10^9)(0.2)}{0.05^2}$

$E = 7.2 \times 10^[11} N/C$

Part b)

Here the point lying of the other side of charge q2

so here electric field due to both charges is in same direction and hence we will add the field due to both charges

So we have

$E = \frac{kq_1}{r_1^2} + \frac{kq_2}{r_2^2}$

$E = \frac{(9\times 10^9)(0.2)}{0.15^2} + \frac{(9\times 10^9)(0.4)}{0.05^2}$

$E = 1.52 \times 10^{12} N/C$

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Topic : electric field intensity

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Answer 2

Answer:

here is your answer

hope this moght help you

Explanation: