2020-04-24T18:19:33-04:00
A charge particle X (-60μC) is on the x axis at the point (2,0) and a particle Y( - 100μC) is on
the x axis at the point (5,0). Calculate
i) The electric field strength and electric potential at the point (4,0). (10 marks)
ii) The coordinates of the point on the x axis where the electric field strength will be zero.
1
2020-04-27T10:15:07-0400
i)
V = k q x r x + k q y r y = − ( 9 ⋅ 1 0 9 ) ( 60 ⋅ 1 0 − 6 2 + 100 ⋅ 1 0 − 6 1 ) V=\frac{kq_x}{r_x}+\frac{kq_y}{r_y}=-(9\cdot10^9)\left(\frac{60\cdot10^{-6}}{2}+\frac{100\cdot10^{-6}}{1}\right) V = r x k q x + r y k q y = − ( 9 ⋅ 1 0 9 ) ( 2 60 ⋅ 1 0 − 6 + 1 100 ⋅ 1 0 − 6 )
V = − 1.17 ⋅ 1 0 6 V V=-1.17\cdot10^{6}V V = − 1.17 ⋅ 1 0 6 V
E = k q x r x 2 − k q y r y 2 = − ( 9 ⋅ 1 0 9 ) ( 60 ⋅ 1 0 − 6 2 2 − 100 ⋅ 1 0 − 6 1 2 ) E=\frac{kq_x}{r_x^2}-\frac{kq_y}{r_y^2}=-(9\cdot10^9)\left(\frac{60\cdot10^{-6}}{2^2}-\frac{100\cdot10^{-6}}{1^2}\right) E = r x 2 k q x − r y 2 k q y = − ( 9 ⋅ 1 0 9 ) ( 2 2 60 ⋅ 1 0 − 6 − 1 2 100 ⋅ 1 0 − 6 )
E = 7.65 ⋅ 1 0 5 V m E=7.65\cdot10^{5}\frac{V}{m} E = 7.65 ⋅ 1 0 5 m V
ii)
k q x r x 2 − k q y r y 2 = 0 → ( 60 ⋅ 1 0 − 6 ( X − 2 ) 2 − 100 ⋅ 1 0 − 6 ( 5 − X ) 2 ) = 0 \frac{kq_x}{r_x^2}-\frac{kq_y}{r_y^2}=0\to\left(\frac{60\cdot10^{-6}}{(X-2)^2}-\frac{100\cdot10^{-6}}{(5-X)^2}\right)=0 r x 2 k q x − r y 2 k q y = 0 → ( ( X − 2 ) 2 60 ⋅ 1 0 − 6 − ( 5 − X ) 2 100 ⋅ 1 0 − 6 ) = 0
X = 3.3 m , Y = 0 m X=3.3\ m, Y=0\ m X = 3.3 m , Y = 0 m
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