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4 charge particles are at the corners of a square of side "a" find q 2q O------------O q 3q O------------O4q E= Ke( q1/r^2)*vector(r)1 + Ke (q2/r^2)* vector r2 + Ke (q3/r^2)*vector r3 E=Ke(2q/a^2)i + ke(3q/2a^2)(icos45.0 + jsin45.0) + ke(4q/a^2)j I gess the question is how they got from...

Final tempeture is:T= 273.15k +75 = 348.15K Final volume: V=9.43L P=1 atm K= 8.20578x10^-2 L*atm* k^-1*mol^-1 PV=nRT n=PV/RT= (1atm)(9.43L)/(8.20578x10^-2 L*atm* k^-1*mol^-1)(348.15K) = 0.330mol Then 0.330mol (16.0g O2/mol) = 5.28g of oxygen scape

Show that the maximum magnitude Emax of the electric field along the azies of a uniformly charged ring occurs at x=a/(2)^(1/2) and has the value Q/(6(3)^(1/2)pieEoa^2) using the fact that: E=KeQx/(x^2 + a^2)^(3/2) The solution manual shows that for maximum , dE/dx=QKe[ (1/(x^2 + a^2)^(3/2) ) -...

3 charged particles are located at the corners of an equilateral triangle calculate the total force y--------------O + 7.00 uC -------O---------------O-----> X +2.00 uC -4.00 uC F1= Ke q1q2/r^2...