This problem emphasizes the fact that even if we have two different charges, mutual electrostatic force between them will be same.
1.8 Coulomb’s Law in Vector Form As force is a vector quantity, it has some magnitude as well as direction. We will write coulomb’s law in vector form so that it will represent magnitude as well as direction of electrostatic force. Consider two charges q1 and q2 separated by distance r. First of all we will define . It is the vector joining charge q1 and q2. Unit vector along Where From Coulomb’s law As direction of force is along Quiz-5 If two charges of magnitude + q1 and – q2 are separated by a distance r, then find out force acting on the charge –q2.
Sol. First define .
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We can see that direction of the force is opposite to that of . This is because the force is attractive in nature.
Illustration 5 A charge of 5 C is placed at the origin. Another charge of – 10C is placed at (3, 4). Find the net force acting on – 10C charge.
Sol.
Here = 5 units As, × ×
Illustration 6 A charge of 10C is placed at (1, 5), another charge of 6C is placed at (4, 2). Find the net force acting on 6C charge.
Sol.
× × 1.9 Electric Field Suppose two charges q1 and q2 are placed at points A and B separated by distance r. Force experienced by a charge at B is given as ... (1) Now suppose there is no charge at point B. If we place a unit positive charge (+1C) at B, it would experience some force. So here at point B we can define electric field as force experienced by +1C charge due to charge q1. Electric field is denoted by E. In above case, we can calculate the value of E at point B due to point charge q1 with this
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Plot a graph between E and x. Sol. Electric field due to a point charge is given as– Direction of electric field to right and left of origin is as shown. Variation of E at different points on x axis will be as shown in the figure below From the above observations, we can plot the graph between E and x. 1.10 Electric Field Lines The concept of electric field lines was first given by Michael Faraday. With the help of this concept of electric field lines, we can visualize electric field more clearly. Suppose we have a charge + Q at point O. We can draw electric field at any point around it, say at point A. If we move further away from the charge +Q along line OA, magnitude of E decreases, so the size of arrow also reduces. We can do the same for all directions. From above diagram, we can observe two things:
1. Direction of electric field - which is along the arrows.
2. Magnitude of electric field - though we are not getting the exact magnitude of E but we can compare the value of E at any two points by comparing the length of the arrows. Faraday joined all these arrows as shown in the figure below and called them electric field
double atan2 (doubly y, double x) It will return the arc tangent in terms of radians of y/x based on the signs of both values to determine the correct quadrant. double cos (double x) It will return the cosine of a radian angle x. double cosh (double x) It will return the hyperbolic cosine of
Using the data provided in each one of these tests it can be assumed that one has done the steps to be able to determine the magnitude and orientation of the charges of the tape in each test, thus, allowing them to apply the same principle to any object they so desired. Their results would line up with the following; that if the two pieces of tape are torn from the same 40 centimeter strip then the tops of both pieces of tape would be positive and the bottoms of both pieces of tape would be negative and that if they would double the tape the attraction or repulsion in general would lower due to the increased density. Their data would also show that two pieces of tape ripped from each other would result in one piece being entirely positive and the other being entirely negative, they would also be able to state that the orientation of how the tape is paired up doesn’t matter.
a). Based on the observation, we assume that the distance between two stations is 0.375 KM Mean time to send = propogation time + transmission time = 375m. + 1000bits 200 x 106 m/sec. 10 000 000 bps. = 102 μsec. b).
-2 -1 -0.50 0.5 1 2 0.5 x 0 0.1 0.1 0.2 0.3 0.2 0.1 5 5 12/9 KING FAHD UNIVERSITY OF PETROLEUM AND MINERALS ACHB/CE EE315: PROBABILISTIC METHODS IN ELECTRICAL ENGINEERING b) M g ( x) x 2 0 x 0 x 2 0 0 x ( a ) b For ( ) ( ) ( ) [ ( )] ( ) ( ) For ( ) ( ) ( ) ( ) ( ) ( ) ( ) { ( ) ( ) ( ) { ∫ ( ) ( ) ∫ ( ) ( ) ( ) 4 4 4 4 13/9 KING FAHD UNIVERSITY OF PETROLEUM AND MINERALS ACHB/CE EE315: PROBABILISTIC METHODS IN ELECTRICAL ENGINEERING 2 G (x) forb x 0 x 0.5 0 x 1 (b) 2 0 x 2 0 x 4 c) Binomial, Poisson (discrete RV), Uniform, Exponential, Rayleigh (Continuous RV) 2 2 2 2 2 14/9 KING FAHD UNIVERSITY OF PETROLEUM AND MINERALS ACHB/CE EE315: PROBABILISTIC METHODS IN ELECTRICAL ENGINEERING QUESTION 3 a) [ ] ∫ ( ) ∫ ∫ ∫ ∫ ([ ] ∫ ) ( [ ] ) =
What are the parts of a rectangle? 4. Ask the students “Do they have a curve? What kind of lines do they have? How many sides does it have?
force field can be detected as it causes " rippling in tiny visible waves" in the air (228). This knowledge eventually leads Katniss to firing an arrow at the arena's force field and breaking out
Each of the three lines share the similarity of rising and to the right in movement. The shape of the 1st line is the most extreme of the three lines rising the fastest. The shape of the second line is less aggressive than the first due to it rising without developing a strong upward curve. The third line is the most consistent of the three rising mostly at an angle with small curvature towards the end. Each line differs the way that they do due to the various amounts of data that creates each line.
((320.5 + 315.8) / 2)) * 100]) and solved to get (1.58%). For the second station we had to determine the distance required to balance the system and the percent difference. To find the unknown distance we set up the equation Fleft*dleft = Fright*dright. We then plugged in the values (11.35 N * x cm = 48cm *
The result of the force acting causes the object 's velocity to either change speed or direction. In conclusion, the impulse experienced by the object equals the change in momentum of the object, which can be seen in equation Ft = m Δ
Newton’s Third Law: - ‘For every reaction, there is an equal and opposite reaction’ - (Hall, 2015). If an object (Object A) exerts a force onto another object (Object B), Object B will exert the same force back, just in the opposite direction. A finger is exerting a force on to the wall. According to Newton’s Third Law of Motion, the wall should be exerting the same amount of force, just in the opposite direction, to the finger.
Now, Cost of equity (Re) = 8.95% + 1.21×7.43% = 17.94% While determining the cost of debt we again used 8.95%,30 year U.S. Government Interest Rate given in Table B as the risk free rate plus 1.10% debt rate premium above Government rate, which is given in Table A. Cost of debt (Rd) = 8.95% + 1.10% =