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mikayleighb2019
18.06.2020 •
Physics
Two charges are located in the xx–yy plane. If q1=−4.10 nCq1=−4.10 nC and is located at (x=0.00 m,y=1.080 m)(x=0.00 m,y=1.080 m), and the second charge has magnitude of q2=3.60 nCq2=3.60 nC and is located at (x=1.20 m,y=0.600 m)(x=1.20 m,y=0.600 m), calculate the xx and yy components, ExEx and EyEy, of the electric field, E⃗ E→, in component form at the origin, (0,0)(0,0). The Coulomb force constant is 1/(4π0)=8.99×109 N⋅m2/C21/(4πϵ0)=8.99×109 N⋅m2/C2.
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Ответ:
Explanation:
Due to first charge , electric field at origin will be oriented towards - ve of y axis.
magnitude
Ey = -8.99 x 10⁹ x 4.1 x 10⁻⁹ / 1.08² j
= - 31.6 j N/C
Due to second charge electric field at origin
= 8.99 x 10⁹ x 3.6 x 10⁻⁹ / 1.2²+ .6²
= 8.99 x 10⁹ x 3.6 x 10⁻⁹ / 1.8
= 18 N/C
It is making angle θ where
Tanθ = .6 / 1.2
= 26.55°
this field in vector form
= - 18 cos 26.55 i - 18 sin26.55 j
= - 16.10 i - 8.04 j
Total field
= - 16.10 i - 8.04 j + ( - 31.6 j )
= -16.1 i - 39.64 j .
Ex = - 16.1 i
Ey = - 39.64 j .
Ответ:
450 cm square
Explanation:
area = length × breadth
A = 30cm × 15cm
A= 450cm square