Asked by Phy
A charge of q1=1×10−6 C is located at corner A of an equilateral triangle. A charge q2=4×10−6 C is located at corner B and a charge of q3=7×10−6 C is located at corner C. The distance AB=0.46 m (the other two sides of the triangle are the same). See the figure.
a) What is the x-component of the Electric field (in V/m) at point P which is located half way between A and C?
b)What is the y-component of the Electric field (in V/m) at point P?
c)What is the Electric Potential (in Volts) at point P?
d)What is the Electric Potential Energy (configuration energy) in Joules of this system of 3 charges?
a) What is the x-component of the Electric field (in V/m) at point P which is located half way between A and C?
b)What is the y-component of the Electric field (in V/m) at point P?
c)What is the Electric Potential (in Volts) at point P?
d)What is the Electric Potential Energy (configuration energy) in Joules of this system of 3 charges?
Answers
Answered by
Shaniqua
Shoot brada brada i got no clue
Answered by
Elena
E =kq/r²
(a) E(x) = E₁-E₃=kq₁/r² - kq₃/r² =
=9•10⁹{(1•10⁻⁶/0.23²)- (7•10⁻⁶/0.23²)}= -1.02•10⁶ V/m
(b) An altitude of a triangle is
h=0.46cos30=0.46•0.866=0.398 m
E(y) = kq ₂/h²= 9•10⁹•4•10⁻⁶/0.398²= - 2.27•10⁵ V/m
(c)
φ=φ₁+ φ₂+ φ₃= kq₁/r + kq₃/h+ kq₃/r=
=9•10⁹{ (1•10⁻⁶/0.23 )+ (4•10⁻⁶/0.398) + (7•10⁻⁶/0.23)} = …
(d)
PE=PE₁₂+PE₁₃+PE₂₃=
= kq₁q₂/r +kq₁q₃/r + kq₂q₃/r=…
(a) E(x) = E₁-E₃=kq₁/r² - kq₃/r² =
=9•10⁹{(1•10⁻⁶/0.23²)- (7•10⁻⁶/0.23²)}= -1.02•10⁶ V/m
(b) An altitude of a triangle is
h=0.46cos30=0.46•0.866=0.398 m
E(y) = kq ₂/h²= 9•10⁹•4•10⁻⁶/0.398²= - 2.27•10⁵ V/m
(c)
φ=φ₁+ φ₂+ φ₃= kq₁/r + kq₃/h+ kq₃/r=
=9•10⁹{ (1•10⁻⁶/0.23 )+ (4•10⁻⁶/0.398) + (7•10⁻⁶/0.23)} = …
(d)
PE=PE₁₂+PE₁₃+PE₂₃=
= kq₁q₂/r +kq₁q₃/r + kq₂q₃/r=…
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