Question

the following situation, Q1 has a charge of 6e and q2 has a charge of e. The two point charges are separated by a distance of r-2.9 m 42 What is the magnitude of the electric field at q2 due to q1? O what is the direction of the electric field at q2 due to qi? ""~ B Soe Incorrect. Tries 1/8 Previous Tries What is the magnitude of the electric field at q1 due to q27 1.71x10 10 N/C what is the direction of the electric กeld at 91 due to q? to the left You are correct. Your receipt no. is 159-8105Previous Tries What is the magnitude of the force on q2 from 1 What is the direction of the force on q2 from q1? st nsr Tries 0/8 If the electric field produced by Q1 at q2 is 2.90x10 10 N/C to the left, what is the magnitude of the force that q, produces on a2 (Fi-2) What is the direction of the force that q1 produces on q2 (Fi-2)? Tries 0/8

Answer 1

(1)
We know that the electric field is given by
$E_{21} = \frac{Kq_{1}}{r^{2}}$
Now putting all the values
$E_{21} = \frac{9*10^{9}*6*1.6*10^{-19}}{2.9^{2}}$
E₂₁ = 10.273*10^-10 N/C
Hence the electric field at the q₂ due to q₁ is 10.273*10^-10 N/C
(2)
Since the charge q₁ is +ve therefore the electric field will be away.
therefore the Electric field will be toward right as shown in the figure.
(3)
Now the electric field at q₁ due to q₂ will be
$E_{12} = \frac{Kq_{2}}{r^{2}}$
$E_{12} = \frac{9*10^{9}*1.6*10^{-19}}{2.9^{2}}$
E₁₂ = 1.712*10^-10 N/C
(4)
Since the charge q₂ is +ve therefore the direction will be away.
hence the direction will be toward left.
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