Asked by tony
Two 2.0 g spheres are suspended by 10.0 cm long light strings. A uniform electric field is applied in the x direction. If the spheres have charges of -5.0 10-8 C and +5.0 10-8 C, determine the electric field intensity that enables the spheres to be in equilibrium at = 10°.
Answers
Answered by
Damon
Mass of sphere = 2*10^-3 kg
Weight of sohere = 9.8*2 *10^-3 = 1.96 *10^-2 N
Force to hold at 10deg = 1.96*10^-2 * tan 10
= 3.45 *10^-3 N
= E * q
so
E = 3.45 *10^-3 / 5 *10^-8
Weight of sohere = 9.8*2 *10^-3 = 1.96 *10^-2 N
Force to hold at 10deg = 1.96*10^-2 * tan 10
= 3.45 *10^-3 N
= E * q
so
E = 3.45 *10^-3 / 5 *10^-8
Answered by
Elise
6.91*10^4 N/C
Answered by
Camille
Follow the same process
F:Static electricity force which the sphere receives
F=qE＋kq^2/r^2
(r=8*10^(-2)×sin11°×2)
sin11°=0.190
the weight of the sphere: mg
(=3*10^(-3)×9.8N)
F/mg=tan11°(=0.194)
F=tan11°*mg
qE－kq^2/r^2= tan11°*mg
E=kq/r^2＋tan11°*mg/q
＝2.90*10^5＋1.90*10^5
=4.80*10^5(N/m)
F:Static electricity force which the sphere receives
F=qE＋kq^2/r^2
(r=8*10^(-2)×sin11°×2)
sin11°=0.190
the weight of the sphere: mg
(=3*10^(-3)×9.8N)
F/mg=tan11°(=0.194)
F=tan11°*mg
qE－kq^2/r^2= tan11°*mg
E=kq/r^2＋tan11°*mg/q
＝2.90*10^5＋1.90*10^5
=4.80*10^5(N/m)
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