document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1185357>Question 1185357</A>: <I><SPAN STYLE=\"word-break: break-all;\"> 1.At a certain point P in an electric field, the magnitude of the electric field is 12 N/C. Calculate the magnitude of the electric force that would be exerted on a point charge of 2.5 x 10-7 C, located at P. \r<BR>\n" );
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document.write( "2.An electron leaves the negative terminal of a cathode ray tube and travels towards the positive plate starting from rest. The electric potential difference between the plates is 2x104 Volts. [mass of electron = 9.11x10-31 kg and the charge on an electron is 1.6x10-19 C]\r<BR>\n" );
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document.write( "(a)Draw a diagram.\r<BR>\n" );
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document.write( "(b)Calculate the electron’s kinetic energy when it reaches the positive plate.\r<BR>\n" );
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document.write( "(c)Calculate the electron’s velocity when it reaches the positive plate.\r<BR>\n" );
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document.write( "3.When a proton is moved against an electric field by an unbalanced force, what happens to the change in electric potential energy? Is it positive or negative? Explain using a diagram and the concept of work.\r<BR>\n" );
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document.write( "4.An electron is travelling horizontally at a speed of 1.5x106 m/s through two horizontal plates. It starts at the negative plate at the top. The plates are 10cm long and the electric field strength between the plates is 100N/C. Calculate the final velocity of the electron as it leaves the plates.<BR>\n" );
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document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #849756 by </B><A HREF=/tutors/aboutme.mpl?userid=CPhill><B>CPhill(1959)</B></A><img src=\"/images/stars/stars-12.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=CPhill><IMG SRC=http://www.algebra.com/images/aboutme-small.gif BORDER=0 alt=\"About Me\" VALIGN=MIDDLE></A>&nbsp;<IMG SRC=http://www.algebra.com/cgi-bin/show-face.mpl?user=CPhill><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=849756\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Here are the solutions to your physics problems:\r<BR>\n" );
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document.write( "**1. Electric Force on a Point Charge:**\r<BR>\n" );
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document.write( "*   **Formula:** Force (F) = Electric Field (E) * Charge (q)<BR>\n" );
document.write( "*   **Calculation:** F = (12 N/C) * (2.5 x 10⁻⁷ C) = 3.0 x 10⁻⁶ N\r<BR>\n" );
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document.write( "**2. Electron in a Cathode Ray Tube:**\r<BR>\n" );
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document.write( "**(a) Diagram:**\r<BR>\n" );
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document.write( "```<BR>\n" );
document.write( "Negative Plate (-)                  Positive Plate (+)<BR>\n" );
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document.write( "|                                                  |<BR>\n" );
document.write( "|  e⁻ (electron) -->                             |<BR>\n" );
document.write( "|                                                  |<BR>\n" );
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document.write( "```\r<BR>\n" );
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document.write( "**(b) Kinetic Energy:**\r<BR>\n" );
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document.write( "*   **Formula:** Kinetic Energy (KE) = Charge (q) * Potential Difference (V)<BR>\n" );
document.write( "*   **Calculation:** KE = (1.6 x 10⁻¹⁹ C) * (2 x 10⁴ V) = 3.2 x 10⁻¹⁵ J\r<BR>\n" );
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document.write( "**(c) Velocity:**\r<BR>\n" );
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document.write( "*   **Formula:** KE = (1/2) * mass (m) * velocity² (v²)<BR>\n" );
document.write( "*   **Rearrange for velocity:** v = √(2 * KE / m)<BR>\n" );
document.write( "*   **Calculation:** v = √(2 * 3.2 x 10⁻¹⁵ J / 9.11 x 10⁻³¹ kg) = √(7.025 x 10¹⁵) ≈ 8.38 x 10⁷ m/s\r<BR>\n" );
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document.write( "**3. Proton Moved Against an Electric Field:**\r<BR>\n" );
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document.write( "*   **Explanation:** When a positive charge (like a proton) is moved *against* an electric field, work is done *on* the charge by the external force. This work increases the electric potential energy of the proton.  The change in electric potential energy is *positive*.\r<BR>\n" );
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document.write( "*   **Diagram:**\r<BR>\n" );
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document.write( "+ (High Potential)<BR>\n" );
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document.write( "|  Force (F) <--- Proton (+)<BR>\n" );
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document.write( "- (Low Potential)<BR>\n" );
document.write( "```\r<BR>\n" );
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document.write( "The electric field points from high potential to low potential. The force on the proton due to the field is in the same direction. To move the proton *against* the field, an external force must act in the opposite direction. This external force does positive work, increasing the proton's potential energy.\r<BR>\n" );
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document.write( "**4. Electron Traveling Through Plates:**\r<BR>\n" );
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document.write( "1.  **Time to traverse plates:**<BR>\n" );
document.write( "    *   **Formula:** time (t) = distance (d) / velocity (v)<BR>\n" );
document.write( "    *   **Calculation:** t = (0.10 m) / (1.5 x 10⁶ m/s) = 6.67 x 10⁻⁸ s\r<BR>\n" );
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document.write( "2.  **Vertical acceleration:**<BR>\n" );
document.write( "    *   **Formula:** Force (F) = Electric Field (E) * Charge (q) = mass (m) * acceleration (a)<BR>\n" );
document.write( "    *   **Rearrange for acceleration:** a = (E * q) / m<BR>\n" );
document.write( "    *   **Calculation:** a = (100 N/C * 1.6 x 10⁻¹⁹ C) / (9.11 x 10⁻³¹ kg) ≈ 1.76 x 10¹³ m/s²\r<BR>\n" );
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document.write( "3.  **Vertical velocity at exit:**<BR>\n" );
document.write( "    *   **Formula:** final velocity (vf) = initial velocity (vi) + acceleration (a) * time (t)<BR>\n" );
document.write( "    *   **Calculation:** vf = 0 + (1.76 x 10¹³ m/s²) * (6.67 x 10⁻⁸ s) ≈ 1.17 x 10⁶ m/s\r<BR>\n" );
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document.write( "4.  **Final velocity (magnitude):**<BR>\n" );
document.write( "    *   The horizontal velocity remains constant at 1.5 x 10⁶ m/s.<BR>\n" );
document.write( "    *   **Calculation:**  Final Velocity = √((1.5 x 10⁶ m/s)² + (1.17 x 10⁶ m/s)²) ≈ 1.9 x 10⁶ m/s\r<BR>\n" );
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document.write( "5. **Final Velocity (direction):**<BR>\n" );
document.write( "    * The angle below the horizontal can be found by taking the inverse tangent of the vertical velocity divided by the horizontal velocity.\r<BR>\n" );
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document.write( "Therefore, the final velocity of the electron as it leaves the plates is approximately 1.9 x 10⁶ m/s.<BR>\n" );
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