$q = \sqrt{\frac{4(0.295(m))^2 (2.20 \times 10^{-4}(kg)) (9.80 \frac{m}{s^2}) (tan(9.40^\circ))}{(9.40^\circ)(sin(9.40^\circ))^2 \times 8.99 \times 10^9 (N) \cdot \frac{m^2}{C^2}}}$
Added by Lourdes R.
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$$q = \sqrt{\frac{4(0.295)^2(2.20 \times 10^{-4})(9.80)(tan(9.40^\circ))(sin(9.40^\circ))^2}{8.99 \times 10^9}}$$ Show more…
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Repeat Problem $24.1$ if the spheres are separated by a center-to-center distance of $1.5 \mathrm{~m}$ in a large vat of water. The dielectric constant of water is about 80 . From Coulomb's Law. $$F_{E}=\frac{k_{0}}{K} \frac{q^{2}}{r^{2}}$$ where $K$, the dielectric constant, is now 80 . Then $$ q=\sqrt{\frac{F_{E} r^{2} K}{k_{0}}}=\sqrt{\frac{(2 \mathrm{~N})(1.5 \mathrm{~m})^{2}(80)}{9 \times 10^{9} \mathrm{~N} \cdot \mathrm{m}^{2} / \mathrm{C}^{2}}}=2 \times 10^{-4} \mathrm{C} $$
Repeat Problem $24.1$ if the spheres are separated by a center-tocenter distance of $1.5 \mathrm{~m}$ in a large vat of water. The dielectric constant of water is about 80 . From Coulomb's Law, $$ F_{E}=\frac{k_{0}}{K} \frac{q^{2}}{r^{2}} $$ where $K$, the dielectric constant, is now 80 . Then $$ q=\sqrt{\frac{F_{E} r^{2} K}{k_{0}}}=\sqrt{\frac{(2 \mathrm{~N})(1.5 \mathrm{~m})^{2}(80)}{9 \times 10^{9} \mathrm{~N} \cdot \mathrm{m}^{2} / \mathrm{C}^{2}}}=2 \times 10^{-4} \mathrm{C} $$
A charge q1 = 1.6 x 10^-19 C (with mass m1 = 9.11 x 10^-31 kg) is 0.09 m away from another charge q2 = 1.6 x 10^-19 C. If q2 is held in place at all times while q1 is released from rest, what is the speed of q1 when it is 0.03 m away from q2? (k = 1/4πε0 = 9 x 10^9 N ∙ m^2/C^2)
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