What is the magnitude of the charge on eachball? (Neglect the mass of the thread
ID: 1664501 • Letter: W
Question
What is the magnitude of the charge on eachball? (Neglect the mass of the thread.) q = CI got the angle A as 12.27 and then
tan12.27 = F/Mg = F/.09g x9.8m/sec^2
.2174 x .09g x 9.8m/sec^2 = F = .1917468
I used F so F=E = kq/r^2
.1917468 = 9.0 x 10^9 x q/.19^2
.1917468 x 19^2/9.0 x 10^9 = 7.69117722 x 10^13 C = q
This the wrong answer. Please help. = 9 * 10 ^ -5 kg Separation of the masses after they come to rest x =19 Distance of the center to the mass afterseparation r = x / 2= 9.5 cm from figure sin = r / L = 13.739 degrees from figure T cos = w = mg from this tension T = mg / cos = [ 9*10 ^ -5 * 9.8 ] / cos 13.739 = 9.265 * 10 ^ -9 N from figure T sin = Fe from this Fe = 2.2 * 10 ^ -9 N we know Fe = K q ^ 2 / x ^ 2 where K = couomb's constant = 8.99 * 10 ^ 9 N m ^2/ C ^ 2 from above charge q = x [Fe / K ] = 0.19 m * 0.4947 * 10 ^ -9 = 0.094 * 10 ^ -9 C Distance of the center to the mass afterseparation r = x / 2= 9.5 cm from figure sin = r / L = 13.739 degrees from figure T cos = w = mg from this tension T = mg / cos = [ 9*10 ^ -5 * 9.8 ] / cos 13.739 = 9.265 * 10 ^ -9 N from figure T sin = Fe from this Fe = 2.2 * 10 ^ -9 N we know Fe = K q ^ 2 / x ^ 2 where K = couomb's constant = 8.99 * 10 ^ 9 N m ^2/ C ^ 2 from above charge q = x [Fe / K ] = 0.19 m * 0.4947 * 10 ^ -9 = 0.094 * 10 ^ -9 C
Explanation / Answer
I ill correct your 2nd answer: You mistake when calculate T T=mg/cos=9,08*10^-4(N) so Fe=Tsin=2,16*10^-4(N) so q = sqrt(Fe*x^2/k)=2,9*10^-8(C)
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