Calculate the amount of each charge in micro Coulombs. Your Answer: 2 equal char

Question

Calculate the amount of each charge in micro Coulombs. Your Answer: 2 equal charges, 194 micro Coulombs each, experiences an attraction force of 194 Newton. Calculate the distance between the charges in mm. Your Answer What is the total charge of all protons in 3 gram of water (H_2O)? Your Answer 3 charges, 5 mu C each, are located on throe vertices A. B. C of an equilateral triangle with sides 1 cm each. Another charge q is located at the mid point D of the side BC. Calculate q in micro Coulomb so that net force on the charge at A due to the charges at B. C and D is zero. Your Answer In a right angle triangle ABC, angle ABC is 90 Degree, AB = 2 m, and angle ACB is 41.81 Degree. A point charge of 5^*8 nC is placed at point C, point charge 4^* 8 nC is placed at point A and point charge 1 C is placed in point B. Calculate the force on charge at B due to others two. Your Answer

Explanation / Answer

Q15) F = kq1q2/r^2 = kq^2/r^2 = 194

9e9*(194*10^-6)^2 / r^2 = 194

r = 1.32 m

Q16) moles of water = n = 3/18 = 0.167

No of water molecules = 6.022*10^23 * 0.167 = 1.00567*10^23 molecules

Actually each molecule of water consists of H2O.

H is just one proton and one electron. No neutrons.
O is 8 protons, 8 neutrons, and 8 electrons.

So one molecule of H2O consists of a total of 10 protons

Hence total proton in 3 gm of water = 1.00567*10^23 * 10 = 1.00567*10^24 protons

Total charge = 1.6*10^-19 * 1.00567*10^24 = 1.609*10^5 C

Q17) Charge q at D must be a negative charge to counter the force due to to B and C

Fab = kQaQb/Rab^2 = 2250 N

Fac = kQaQc/Rac^2 = 2250 N

Rad = sqrt(1^2 - 0.5^2) = 0.866 cm

Fad = kQaQd/Rad^2 = 6*10^8 * q

For the total force to be zero

Fad = (Fab+Fac)*cos(30)

6*10^8 * q = 3897.11

q = 6.5*10^-6 C = 6.5 uC

Q18) Let point B be the origin, then point A will be on positive y axis and C on positive x axis

Let sides of the triangle ABC be a,b and c

a is lenght of AB, b is lenght of BC, c is lenght of AC

Then using sine rule

a/sinA = b/sinB

A = 90 - 41.81 = 48.19 deg

B = 90 deg

a = AB = 2m

b = BC = asinA/sinB = 1.49 m

Fba = kQaQb/Rab^2 = 9e9*4.8e-9*1 / 2^2 = 10.8 N in negative y direction

Fbc =  kQcQb/Rbc^2 = 9e9*5.8e-9*1 / 1.49^2 = 23.5 N in in negative x direction

Total Force on B = sqrt(10.8^2 + 23.5^2) = 25.86 N

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