Three charges arrange in the form of an equilateral triangle.?
How EXACTLY do you expect us to draw forces on Q3 Other questions: F = kQq / r^2 Plug and chug
Three charges are arranged on the vertices of an isosceles triangle as shown in the diagram.?
Can you please show the diagram?
The three charges shown in the figure are located at the vertices of an isosceles triangle.?
Calculate the electric potential at the mid-point of the base if the magnitude of the positive charge is 7.0×10−9 C and the magni- tude of the negative charges are 5.9 × 10−9 C. The value of the Coulomb constant is 8.98756 × 109 N · m2/C2 , and the accelera- tion of gravity is 9.8 m/s2 . Answer in units of V the two sides of the triangle are 5.4cm and the base is 1.4cm Please help? due today at midnight. I don't know where to start or how to solve this problem if you could please help!..i know Q+ is 7.0 x 10-9,Q- is 5.9 X 10-9,k =8.98755 x10^9,g= 9.8m/s. im not able to solve this equation
The three charges in Figure P16.17 are at the vertices of an isosceles triangle. Calculate the electric potent?
Terribly sorry. The three charges in Figure P16.17 are at the vertices of an isosceles triangle. Calculate the electric potential at the midpoint of the base, taking q = 5.00 nC. Here x = 1.40 cm. The figure wont load but its an Isosceles triangle with the two congruent sides being 4 cm. The midpoint is on the in congruent side and the q charges on either side of it are -. The charge q at the top point of the triangle is positive.
Three charged particles are placed at the corners of an equilateral triangle of side d = 1.30 m (Fig. 16-53).?
Q1) Magnitude of force on Q1 due to Q3 = F₁₃ = k*Q1*Q3/1.3² Magnitude of force on Q1 due to Q2 = F₁₂ = k*Q2*Q1/1.3² Now you must add up the components in vector form: Fx = F₁₂cos(60°) - F₁₃cos(60°) Fy = F₁₂sin(60°) + F₁₃sin(60°) 60° because it's an equilateral triangle. You'll notice the signs of the forces - both F₁₂ and F₁₃ are negative, and only the x-component of the F₁₃ force is positive. Q2) F₂₁ = k*Q2*Q1/1.3² F₂₃ = k*Q2*Q3/1.3² Fx = -F₂₁cos(60°) - F₂₃ Fy = -F₂₁sin(60°) F₂₁ is negative, which means it's an attractive force, so Q1 is pulling Q2 in the positive x-direction and the positive y-direction. F₂₃ is positive, which means it's repulsive, so Q3 is pushing Q2 in the negative x-direction. Q3) F₃₁ = k*Q3*Q1/1.3² F₃₂ = k*Q3*Q2/1.3² Fx = F₃₁cos(60°) + F₃₂ Fy = -F₃₁sin(60°) F₃₁ is negative, and Q1 pulls Q3 in the negative x-direction, but the positive y-direction. F₃₂ is positive, and Q2 pushes Q3 in the positive x-direction. To determine magnitude of the net force on a particle, F = sqrt(Fx²+Fy²), while direction is θ=arctan(Fy/Fx). Make sure you convert µC into Coulombs before you get started.
Consider the charges q,q and -q placed at the vertices of an equilateral triangle of each side l. What is force on each charge?Show that F1+ F2+ F3=0.
By using coloumbs law of charge you easily found the force experience by each particle and you easily prove that the vectorial sum of forces on each charges is ‘0’.But,you were thinking that the resultant forces get ‘0’ due to equilateral arrangement of charges,No that's not,in any arrangements of charges you will get '0′ resultant forces,that sounds very interesting.Means if you have infinite number of charges (of whatever magnitude or sign) arrange in any configurations you will find that the resultant force is '0′.The resultant force is zero everytime because of Newton third law of motion:- “every action is equal and opposite reaction”Due to Newton's third law all forces get cancelled(as you see in the image)I hope this helps you,in case of any doubt comment below
Two equal charges each of q are placed at the vertices A and B of an equilateral triangle of length a. What would be the intensity of electric field at vertice C?
Two equal charges [math]q[/math] are placed on two vertices [math]A[/math] and [math]B[/math] of an equilateral triangle of side length [math]a[/math]. We want to determine the intensity of the electric field at the third vertex [math]C.[/math]Let us assume that the charges are positive.The intensity of the electric field due to the charge at vertex [math]A[/math] is [math]E_A = \frac{1}{4\pi \epsilon_0k}.\frac{q}{a^2}[/math] in the direction [math]AC.[/math]The intensity of the electric field due to the charge at vertex [math]B[/math] is [math]E_B = \frac{1}{4\pi \epsilon_0k}.\frac{q}{a^2}[/math] in the direction [math]BC.[/math]For air or vacuum [math]k=1.[/math]The angle between [math]AC[/math] and [math]BC[/math] is [math]60^o.[/math][math]\Rightarrow \qquad[/math] The net electric field would be [math]\vec E = \vec {E_A}+\vec {E_B}.[/math]The magnitude of [math]\vec E = |\vec E| = 2.\frac{1}{4\pi \epsilon_0k}.\frac{q}{a^2}.\cos 30^o = \frac{\sqrt 3}{4\pi \epsilon_0k}.\frac{q}{a^2}.[/math]The direction of [math]\vec E[/math] is perpendicular to [math]AB[/math] in the direction from line [math]AB[/math] to point [math]C.[/math]In case the charges are negative, the magnitude of the net electric field would be the same but the direction would be perpendicular to [math]AB[/math] in the direction from point [math]C[/math] to line [math]AB.[/math]
Three charges, each of the q coulomb, are kept at the vertices of an equilateral triangle of side metre. What is the resultant force acting on any one of the charges, and the potential energy of the system?
This question is challenging to answer without a diagram. However drawing one is labor intensive and yet a necessity to fire up the imagination. I have taken the time to sketch one.Pairs of charges exert forces on each other. In the arrangement that you described, one charge is repelled by two charges. We find the resultant force on say, the uppermost charge by the principle of superposition of the effective components of each force.Because of the symmetry of an equilateral triangle and equality of charges, [math]q[/math], the leftward and rightward components of force on the upper charge will add to zero. Thus they do not contribute to the resultant force, [math]R[/math].Resultant ForceThe resultant of the vertical components of each of the forces is given by their vector sum.It can be evaluated by the cosine rule. For this we need angle [math]M[/math] which can be shown to be [math]120^\circ[/math] since each angle of an equilateral triangle is [math]60^\circ[/math].[math]R=F_{qqy}^2+F_{qqy}^2-2F_{qqy}F_{qqy}cos120^\circ[/math][math]R=2F_{qqy}^2-2F_{qqy}^2cos120^\circ[/math][math]R=2F_{qqy}^2[1-(-0.5)][/math][math]R=2F_{qqy}^2(1+0.5)[/math][math]R=3F_{qqy}^2 \tag 1[/math]But [math]F_{qqy}=\frac{q\times q}{4\pi\epsilon_0 r^2}[/math]where [math]\epsilon[/math] is the permittivity of fee space and[math] r=1m[/math] is the separation between charges.[math]F_{qqy}=\frac{q^2}{4\pi\epsilon_0\times 1^2}[/math][math]F_{qqy}=\frac{q^2}{4\pi\epsilon_0}[/math]∴ equation [math](1)[/math] becomes [math]R=3\left(\frac{q^2}{4\pi\epsilon_0}\right)^2[/math]Potential energyThe potential energy in the system is a scalar quantity so we find the potential energy between each pair of charges and then add them up.For a pair of charges q and q[math]E_p=-\frac{q^2}{4\pi\epsilon_0r}[/math] and since [math]r=1m[/math]For three pairs of charges; [math]\displaystyle E_{p-total}=-3\frac{q^2}{4\pi\epsilon_0}[/math]The negative shows that energy is spent by the electric fields to cause repulsion.
ABC is an equilateral triangle. Charges +q are placed at each corner. What is the electric intensity at O?
May I advise you young students to not to ask text book questions if you are a serious physics .Do not aspire merely to see your name in print in Quora by asking simple questions. I am sure you are JEEE aspirants and therefore tackle typical and conceptual questions to sharpen your skill. If you like my suggestion do shoot back!
Three charges -q, Q and -q are placed respectively at equal distances on a straight line. If the potential energy of the system of three charges is zero, then what is the ratio of Q:q?
Let us assume the distance between the charges is 'd' and let 'e' denote permittivity in vacuum since I cannot type the symbol, then the total potential of the system will be([math]1/4πe)*(-qQ)/d + [/math]([math]1/4πe[/math])[math]*(Q(-q))/d + (1/4πe)*(-q*-q)/2d = 0[/math]Solving the above will give you the ratio Q:q=1:4