Answer:
3, 5, 7, 9, 11, .........
Step-by-step explanation:
Given
= n(n + 2) , then
S₁ = 1(1 + 2) = 1(3) = 3 ⇒ a₁ = 3
S₂ = 2(2 + 2) = 2(4) = 8
S₃ = 3(3 + 2) = 3(5) = 15
Thus
a₂ = S₂ - S₁ = 8 - 3 = 5
a₃ = S₃ - S₂ = 15 - 8 = 7
The first 3 terms are 3, 5, 7
This is an AP with common difference d = 2, then
a₄ = a₃ + 2 = 7 + 2 = 9
a₅ = a₄ + 2 = 9 + 2 = 11
and so on
The electric field strength at any point from a charged particle is given by E = kq/r^2 and we can use this to calculate the field strength of the two fields individually at the midpoint.
The field strength at midway (r = 0.171/2 = 0.0885 m) for particle 1 is E = (8.99x10^9)(-1* 10^-7)/(0.0885)^2 = -7.041 N/C and the field strength at midway for particle 2 is E = (8.99x10^9)(5.98* 10^-7)/(0.0935)^2 = <span>-7.041 N/C
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Note the sign of the field for particle 1 is negative so this is attractive for a test charge whereas for particle 2 it is positive therefore their equal magnitudes will add to give the magnitude of the net field, 2*<span>7.041 N/C </span>= 14.082 N/C
Answer:
6th term
Step-by-step explanation:
95=55+8n-8
95-55=8n-8
40+8=8n
48/8=8n/8
6=n
