Answer to Question #295527 in Electric Circuits for Michael

Question

Suppose you have 5 point charges q1 to q5 distributed along a semi-circle maintaining equal distance. The charge q6 is on the center of the circle which is also the origin of given coordinate system. The radius of the semi-circle is R=15cm and the magnitude of the charges are as follows : q1=q5=29μC , q2=q4=36μC and q3=q6=−37μC .

Find the x and y component of the vector for all the following questions below

a) Find the vector that points from q1 to q6 and call it r(1,6) .

b) Calculate the coulomb force on q6 due to q1 in unit vector notation and call it F(1,6).

c) Find the vector that points from q3 to q6 and call it r(3,6).

d) Calculate the coulomb force on q6 due to q3 in unit vector notation and call it F (3,6).

e) Find the vector that points from q4 to q6 and call it r(4,6).

f) Calculate the coulomb force on q6 due to q4 in unit vector notation and call it F (4,6).

g) Calculate the Net Electric force on q6 due to all other charges.

Answer 1

a)

The distance between the charge $q_1$ and $q_6$ is $15cm=0.15m$

The vector points in positive x-axis.

Therefore

$r_{(1,6)}=0.15i$

b)

The force is attractive toward negative x-axis

$F=\frac{k{(q_1,q_2})}{r^2}$

$F=\frac{9×10^9(37×29×10^{-12})}{0.15^2}=429.2N$

$F_{(1,6)}=-429.2i$

c)

The distance from $q_3\>to \>q_6$

$=15cm= 0.15m$

The vector points in negative y-axis

$\therefore r_{(3,6)}=-0.15j$

d)

The force is repulsive toward negative y-axis,

$|F|=\frac{9×10^9(37×37×10^{-12})}{0.15^2}=547.6N$

$F_{(3,6)}=-547.6j$

e)

The angle between the vector and x-axis

$=\frac{180}{4}=45°$

The vector is sum of y-component and x- component

$\therefore r_{(4,6)}=-0.15\>Cos\>45(i)+-0.15\>Sin\>45j$

$=-0.11i-0.11j$

f)

The force is attractive toward $q_4$ with x- and y- components equal.

This is because $sin45= cos45$

$F=\frac{9×10^9(36×37×10^{-12})}{0.15^2}=532.8$

$532.8\>Sin\>45=376.75$

$F_{(4,6)}=376.75i+376.75j$

g)

Net force on $q_6$ due to $q_1$ and $q_{5}=0$

Also net force due to x-component of $F_{4,6}\>and\>F_{2,6}$ $=0$

They point in opposite direction.

$F_{(4,6)}=F_{(2,6)}$

Net force $=(0)i+[2(376.75)+(-547.6)]j$

$=205.9j$