Asked by Stephanie6
In Fig. 21-24, the particles have charges q1 = -q2 = 400 nC and q3 = -q4 = 92 nC, and distance a = 5.0 cm. What are the (a) x and (b) y components of the net electrostatic force on particle 3?
The figure cannot be uploaded, but the particles are located as a square, with q1 in the upper left, q2 in the upper right, q3 in the lower left, and q4 in the lower right.
The figure cannot be uploaded, but the particles are located as a square, with q1 in the upper left, q2 in the upper right, q3 in the lower left, and q4 in the lower right.
Answers
Answered by
Elena
F13=k•q1•q3/a² =
=9•10^9•400•10^-9•92•10^-9/25•10^-4=
=0.132 N
F13(x) =0, F13(y) =0.132 N
F43= k•q4•q3/a² =
=9•10^9•92•10^-9•92•10^-9/25•10^-4=
=0.0305 N
F43(x)= - 0.0305 N. F43(y)= 0
F23= k•q2•q3/(a√2)²= 9•10^9•400•10^-9•92•10^-9/1.41•25•10^-4=0.0936 N
F23 directed along the diagonal of the square (towards q2)
F23(x) =F23(y)=
=0.0936•cos 45=0.0936•0.707=0.0662 N.
F(x) = F13(x) +F43(x) +F23(x) =
=0 - 0.0305+ 0.0662= 0.0357 N,
F(y) = F13(y) +F43(y) +F23(y)= 0.132+0+0.0662 =0.1982 N.
=9•10^9•400•10^-9•92•10^-9/25•10^-4=
=0.132 N
F13(x) =0, F13(y) =0.132 N
F43= k•q4•q3/a² =
=9•10^9•92•10^-9•92•10^-9/25•10^-4=
=0.0305 N
F43(x)= - 0.0305 N. F43(y)= 0
F23= k•q2•q3/(a√2)²= 9•10^9•400•10^-9•92•10^-9/1.41•25•10^-4=0.0936 N
F23 directed along the diagonal of the square (towards q2)
F23(x) =F23(y)=
=0.0936•cos 45=0.0936•0.707=0.0662 N.
F(x) = F13(x) +F43(x) +F23(x) =
=0 - 0.0305+ 0.0662= 0.0357 N,
F(y) = F13(y) +F43(y) +F23(y)= 0.132+0+0.0662 =0.1982 N.
Answered by
Ivan
This answers is wrong don't use please.
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