**Problem Statement:**A small object is depicted with a mass of 1 gram and a charge of -3 nC. When an electric field \( E \) is turned on, the object moves into equilibrium. The diagram shows a tension \( T \) of 10 mN in the string, and it makes an angle of 30 degrees with the horizontal axis.**Objective:**Find the magnitude and direction of the electric field \( E \).**Options:**a. \(|E| = 5.7 \times 10^4\) N/C at 60 degrees; 3rd Quadrant (240 degrees)b. \(|E| = 3.3 \times 10^4\) N/C at 30 degrees; 3rd Quadrant (210 degrees)c. \(|E| = 5.7 \times 10^4\) N/C at 30 degrees; 1st Quadrant (30 degrees)d. \(|E| = 3.3 \times 10^4\) N/C at 90 degrees; 1st Quadrant (60 degrees)**Diagram Explanation:**The diagram illustrates an object suspended by a string at an angle of 30 degrees from the horizontal axis. The tension in the string (T) is indicated as 10 mN. The object is subject to its gravitational force, the electric force due to its charge, and the tension in the string. The task is to determine how these forces interact to achieve equilibrium.The tension vector is at a 30-degree angle above the horizontal, indicating the need for resolving forces along the horizontal (x) and vertical (y) axes. The electric field \( E \) mentioned will counteract these forces, and its values and directions need to be calculated accordingly.**Note:**To solve this problem, one would use equilibrium equations considering both the forces due to the electric field and the gravitational force acting on the charged object.

S. The pith ball shown below has a mass of I gram and a charge of-3oC. When an electric field, E, is tumed on the pih ball moves into equilibrium as shown below. The tension in the string is 10mN at 30 to the horizontal. m1 gm T 10mN Find the magnitude and the direction of the E field a. JE- 5.7*10 NC @ 60 deg 3rd Quadrant (240 deg) b. JE-33*10 Nc @ 30 deg 3rd Quadrant (210 deg) c. E-5.7*10 NC @ 30 deg Ist Quadrant (30 deg) d. JE-33*10 N/C @ 60 deg Ist Quadrant (60 deg)

S. The pith ball shown below has a mass of I gram and a charge of-3oC. When an electric field, E, is tumed on the pih ball moves into equilibrium as shown below. The tension in the string is 10mN at 30 to the horizontal. m1 gm T 10mN Find the magnitude and the direction of the E field a. JE- 5.710 NC @ 60 deg 3rd Quadrant (240 deg) b. JE-3310 Nc @ 30 deg 3rd Quadrant (210 deg) c. E-5.710 NC @ 30 deg Ist Quadrant (30 deg) d. JE-3310 N/C @ 60 deg Ist Quadrant (60 deg)

College Physics

11th Edition

ISBN:9781305952300

Author:Raymond A. Serway, Chris Vuille

Publisher:Raymond A. Serway, Chris Vuille

Chapter1: Units, Trigonometry. And Vectors

Section: Chapter Questions

Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...

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Question

**Problem Statement:**

A small object is depicted with a mass of 1 gram and a charge of -3 nC. When an electric field \( E \) is turned on, the object moves into equilibrium. The diagram shows a tension \( T \) of 10 mN in the string, and it makes an angle of 30 degrees with the horizontal axis.

**Objective:**

Find the magnitude and direction of the electric field \( E \).

**Options:**

a. \(|E| = 5.7 \times 10^4\) N/C at 60 degrees; 3rd Quadrant (240 degrees)

b. \(|E| = 3.3 \times 10^4\) N/C at 30 degrees; 3rd Quadrant (210 degrees)

c. \(|E| = 5.7 \times 10^4\) N/C at 30 degrees; 1st Quadrant (30 degrees)

d. \(|E| = 3.3 \times 10^4\) N/C at 90 degrees; 1st Quadrant (60 degrees)

**Diagram Explanation:**

The diagram illustrates an object suspended by a string at an angle of 30 degrees from the horizontal axis. The tension in the string (T) is indicated as 10 mN. The object is subject to its gravitational force, the electric force due to its charge, and the tension in the string. The task is to determine how these forces interact to achieve equilibrium.

The tension vector is at a 30-degree angle above the horizontal, indicating the need for resolving forces along the horizontal (x) and vertical (y) axes. The electric field \( E \) mentioned will counteract these forces, and its values and directions need to be calculated accordingly.

**Note:**

To solve this problem, one would use equilibrium equations considering both the forces due to the electric field and the gravitational force acting on the charged object.

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