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homework_02_electric_field_from_arc_of_charge
A total charge [Q = -2.0 \mu C] is distributed uniformly over a quarter circle arc of radius [a = .095 m] as shown.
What is [\lambda] the linear charge density along the arc?
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[2 \pi r]
- Circumfrerence
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Let
- [ r = a = .095 m]
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[\lambda = \frac{ k \frac{ Q}{ r^2}}{ \frac{ 1}{ 2} \pi r}]
- It says I made a power of ten error, what I am doing wrong?
What is [E_x], the value of the [x]-component of the electric field at the origin [(x, y) = (0,0)] ?
need to use [\vec E = \int{ \frac{ 1}{ 4 \pi \varepsilon_0} \frac{ \hat r}{ r^2} dq}] and convert it to polar coodinates [\lambda r f \theta]
[\int{ \frac{ k}{ a^2} \hat r d q} = \int\limits_0^{\frac{ \pi}{ 2}}{ \frac{ k}{ a^2} \left( -\hat x \cos \theta, -\hat y \sin \theta\right) \frac{ 2q}{\pi a} a d \theta} = \frac{ k}{ a^2} \frac{ 2q}{ \pi}( -\hat x -\hat y) = \frac{ k}{ a^2} \frac{ 2^{\frac{ 3}{ 2}} q}{ \pi} \left( \frac{ -\hat x}{ \sqrt{ 2}}, \frac{ -\hat y}{ \sqrt{ 2}}\right)]
What is [E_y], the value of the [y]-component of the electric field at the origin [(x,y) = (0,0)]?
How does the magnitude of the electric field at the origin for the quarter-circle arc you have just calculated comnpare to the electric field at the origin produced by a point charge [Q = -2.0 \mu C] located a distance [a = 0.095 m] from the origin along a [45^\circ] line as shown in the figure?
- The magnitude of the field from the point charge is greater than
that from the quarter-arc of charge
- The field from the point charge is just given by [E_{point} = \frac{ k Q}{a^2}]. You can find the magnitude of the field from the arc of charge from you answers to b) and c). Namely, [E_{arc} = \frac{ k Q}{ a^2} \frac{ 2 \sqrt{2}}{ \pi}]. Therefore, the field from the point charge is larger than the field from the arc by a factor of [ \frac{ \pi}{ 2 \sqrt{2}}].
What is the magnitude of the electric field at the origin produced by a semi-circular arc of charge [= -4.0 \mu C], twice the charge of the quarter-circle arc?