Consider the figure below. 1. Find the electric field (in N/C) at x = 6.00 cm in

Question

Consider the figure below.

1. Find the electric field (in N/C) at x = 6.00 cm in part (a) of the figure, given that q = 5.00 µC. (Indicate the direction with the sign of your answer.) ?

2. For part (a) of the figure, at what position (in cm) between 3.00 and 8.00 cm is the total electric field the same as that for 2q alone?

3. For part (a) of the figure, at what position (in cm) to the right of 11.0 cm is the total electric field zero, other than at infinity?

Explanation / Answer

we know electric field is given as

E=kq/(r^2)

1)Field at 6 cm

E=9*10^9*[{5*10^{-6}/(.03^2)}+{-2*5*10^{-6}/(.02^2)}+{5*10^{-6}/(.05^2)}]=9*10^9*10^{-10}*-1.74=-1.57N/C

2)For the total electric field to be same as that of -2q the point should be exactly between thees two ,i.e.,at x= 7cm

3)Let at x cm from 11cm the electric field be zero so

0=9*10^9*[{5*10^{-6}/((.08+x)^2)}+{-2*5*10^{-6}/((.03+x)^2)}+{5*10^{-6}/((x)^2)}]

0=[{5*10^{-6}/((.08+x)^2)}+{-2*5*10^{-6}/((.03+x)^2)}+{5*10^{-6}/((x)^2)}]

{-2*5*10^{-6}/((.03+x)^2)}=[{5*10^{-6}/((.08+x)^2)}+{5*10^{-6}/((x)^2)}]

{2*5}/((.03+x)^2)}=[{5/((.08+x)^2)}+{5/((x)^2)}]

x=.196 m=19.6 cm

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