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Consider a plastic ring of radius R = 50.0 cm on which there are two charged beads as shown in the figure above.

Bead 1 has q1 = +2.00 μC and is fixed in place on the x-axis. Bead 2 has q2 = +6.0 μC and can be moved along the ring.

a) Determine the positive angle q for q2 which can produce a net Electric field of magnitude E = 2.00 x 10^5 N/C

at the centre of the ring ??

b) Is there any negative angle q’ for q2 which can produce a net Electric field of magnitude E = 2.00 x 10^5 N/C at

the centre of the ring ?? If so, calculate that angle.

Sagot :

Okpalawalter8idn

The positive angle and negative angle is mathematically given as

[tex]\theta=66.8[/tex]  counterclockwise from the x-axis

[tex]\theta=293.28[/tex] counterclockwise from x-axis

What are the positive angle and negative angle which can produce a net Electric field of magnitude E = 2.00 x 10^5 N/C ?

Question Parameters:

R = 50.0 cm

Bead 1 has q1 = +2.00 μC and is fixed in place on the x-axis.

Bead 2 has q2 = +6.0 μC and can be moved along the ring.

E = 2.00 x 10^5 N/C

Generally, the equation for the  Electric field  is mathematically given as

[tex]E=\frac{kq1}{R^2}\\\\Therefore\\\\E1=\frac{9*10^9*2*10^6}{0.5^2}\\\\E1=7.49*10^4N/C[/tex]

Where

[tex]E=\frac{kq1}{R^2}*(cos(i)-sin(j)\\\\E1=\frac{9*6*10^6}{0.5^2}*(cos(i)-sin(j)\\\\\E1=2.17*10^5*(cos(i)-sin(j)\\\\[/tex]

Hence

[tex]E=\sqrt{(0.75-2.17*cos\theta)^2+(2.174sin2\theta)} *10^5=2*10^5\\\\\5.2888-3.261cos\theta=4\\\\Cos\theta=0.3952\\\\[/tex]

[tex]\theta=66.8[/tex]  counterclockwise from the x-axis

b)

for the negative angle

[tex]Cos\theta=cos66.8\\\\Cos \theta=cos(360-66.72)[/tex]

[tex]\theta=293.28[/tex] counterclockwise from x-axis

For more information on Electric field

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