Question

q1,q2,q3 charges placed on a bisector triangle in a straight line q1,q3 at edges and q2 at center P is a point placed at top of the triangle. q1=q2=-q3=2uc. distance between q1q2=q2q3=2cm,pq1=pq3= 4cm.determine the magnitude and the direction of the electric field at point P.

Answer 1

Answer on Question #69554, Physics / Electromagnetism

Question: q1,q2,q3 charges placed on a bisector triangle in a straight line q1,q3 at edges and q2 at center P is a point placed at top of the triangle. q1=q2=-q3=2uc. distance between q1q2=q2q3=2cm,pq1=pq3=4cm.determine the magnitude and the direction of the electric field at point P.

Solution:

$$\vec {E} = \vec {E} _ {1} + \vec {E} _ {2} + \vec {E} _ {3}.$$ $$E = E _ {1} + 2 E _ {2} \cos \alpha = \frac {k q}{r _ {1} ^ {2}} + \frac {2 k q}{r _ {2} ^ {2}} \frac {\sqrt {1 2}}{2} = 9 * 1 0 ^ {9} * 2 * \frac {1 0 ^ {- 6}}{4 * 1 0 ^ {- 4}} + 9 * 1 0 ^ {9} * 4 * \frac {1 0 ^ {- 6}}{1 6 * 1 0 ^ {- 4}} \frac {\sqrt {1 2}}{2} =$$ $$(4. 5 + 3. 8 9) * 1 0 ^ {7} = 8. 3 9 * 1 0 ^ {7} \frac {N}{C} (u p w a r d s)$$

Answer: $8.39 * 10^{7} \frac{N}{C}$ (upwards)

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