Consider a parallel universe where the laws of nature as we know them are differ

Question

Consider a parallel universe where the laws of nature as we know them are different, and aqueous solutions can have an excess charge. Two beakers are separated 1.90 meters and each contains 0.500L of an aqueous solution of 6.20x10^-6 M excess charge. One beaker has a net positive charge, and one beaker has a net negative charge. Calculate the force between the beakers using the formula below.

Map sapling learning Consider a parallel universe where the laws of nature as we know them are different, and aqueous solutions can have an excess charge. Two beakers are separated 1.90 meters and each contains 0.500 L of an aqueous solution of 6.20 x 106 M excess charge. One beaker has a net positive charge, and one beaker has a net negative charge. Calculate the force between the beakers using the formula below, 1 4142 F= where F is force, eo is the permittivity constant and is equal to 8.85% 10% C2/(N-m2). q1 is the charge contained in the first beaker, q2 is the charge contained in the second beaker, and r is the distance between the beakers. Note that 1 mole of charge is 96,485 C, and because one of the beakers contains a net negative charge, one q will be negative, and F will be negative. A negative F corresponds to an attractive force. Number F=

Explanation / Answer

moles of charge in a beaker, q1 = q2 = molar concentration*volume of solution in litres = 6.2*10-6*0.5 = 3.1*10-6 moles

Thus, charge carried by the beakers = moles*Faraday's constant = 3.1*10-6 * 96485 = 0.3 C

Thus, force F = (0.3)*(-0.3)/(4*3.14*8.85*10-12*1.92) = 2.24*108 N

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