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q02b
Kurt Robert Rudolph edited this page Jun 13, 2012
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- [F_{ij} = \kappa \frac{ q_i q_j}{ r^2} \hat r]
- [ \kappa = \frac{ 1}{ 4 \pi \epsilon_0} = 8.99 \times 10^9 \frac{ Nm^2}{ C^2} ]
- [ \mu = 10^{-6}]
- Let:
- [ a ] be the height of the rectangle
- [ 3a ] be the lenght of the rectangle
- [ d = \sqrt{ a^2 + (3a)^2 }]
- [ \theta = \arctan{ \left( \frac{ \frac{ a}{ 2}}{ \frac{ 3a}{ 2}}\right)}]
- [ F_Q = \kappa \frac{ q Q}{ \left(\frac{ d}{ 2} \right)^2} ]
- [ \sin{ (\theta)} = \frac{ \frac{ a}{ 2}}{ \frac{ d}{ 2}} ]
- [ \cos{ (\theta)} = \frac{ \frac{ 3a}{ 2}}{ \frac{ d}{ 2}} ]
- [ \tan{ (\theta)} = \frac{ \frac{ a}{ 2}}{ \frac{ 3a}{ 2}} ]
- [ F_{Q,x} = \kappa \frac{ q}{ \frac{ d}{ 2}} \cos{ (\theta)} ]
- [ F_{Q,y} = \kappa \frac{ q}{ \frac{ d}{ 2}} \sin{ (\theta)} ]
- Let:
- [ q = 3 \mu C ]
- [ a = .02 m ]
- Then:
- [ d = \sqrt{ a^2 + (3a)^2 } = 0.063246]
- [ F_Q = \kappa \frac{ q}{ \left(\frac{ d}{ 2} \right)^2} = 2.697 \times 10^7 ]
- Let:
- [ q = 3 \mu C ]
- [ Q = 4 \mu C ]
- [ a = .02 m ]
- Then:
- [ d = \sqrt{ a^2 + (3a)^2 } = 0.063246]
- [ F_Q = \kappa \frac{ q Q}{ \left(\frac{ d}{ 2} \right)^2} = 107.88 ]